Challenging Mathematical Teasers

CHALLENGING MATHEMATICAL TEASERS J. A. H. HUNTER

DOVER PUBLICATIONS, INC. NEW YORK

Copyright © 1980 by J. A. H. Hunter. All rights reserved under Pan American and International Copyright Conventions. Published in Canada by General Publishing Company, Ltd.,

30 Lesmill Road, Don Mills, Toronto, Ontario.

Published in the United Kingdom by Constablt' and Com­ pany, Ltd., IO Orange Street, London WC2H 7EG.

Challenging Mathematical Teasers

is a new work, first

published by Dover Publications, Inc., in 1980.

International Standard Book Number: 0-486-2.38.S2-0 Library of Congress Catalog Card Number: 79-.)1888 Manufal'lured in the Unitt'd Statl's of Am('rica Dover Publications, 1m'.

180 Varil-k Strl'et

New York. N.Y. )()014

PREFACE I n previous collections of my problems the more diffi­ cult h ave been avoi d e d . But there seems to be a growing d e m and for problems at a somewhat more challe ngi ng level. The two Appe ndices provi de bri ef outlines of concepts that may not be famili ar to some readers, concepts that are de alt with m ore fully in textbooks . Once again I must thank the thousands of kin d news­ paper and magazine readers whose encou rage ment, loyal support, and even i d e as h ave been l argely re sponsible for publication of this little book. H ave fu n ! J . A . H . HUNTER

January 1979

CONTENTS

TE ASERS

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24.

Pecki ng Order F o r the One That G o t Away Long Odds Like F ather Like Son W h o Spilled t h e Coffee? So F a r Apart All Twos Those Share s You H ave t o Know How S avings A R andom Mailing No Computer for This At the Diner They Don't C o m e Singly Who Glubs Glygons? Three i n a Row A Switch J u st Tri angles An Ancient Problem Similar but Differe nt J u st for Kids The Pati o Square s and Squ are s Wrong but Right

Problem Page

1 1 2 2 2 3 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 7 8 8

Solutiol Page

37 37 38 38 39 40 40 41 41 41 42 42 43 43 43 44 44 45 45 46 46 47 47 48

2 5 . T h e Stamp 26. A Tale of Two Guys 27. Just Junk 28 . A Simple Routine 29. Comme moration Stamps 30. Almost Neighbors 3 1 . Real E state 3 2 . Greetings 33. Away from It All 34. Double the Odds 3 5 . B ounty Hunters 36. The Mural 37. A Woman's J ob 38. E ach to E ac h 39. The J ewel Box 4 0 . Peter's Pe nnies 4 1 . S o Very Simple 42. Power Play 43. The Collector 44. A M atter of Squ are s 45. The Only Tree 46. I t M ake s Sense 47. Grandad's B irthday 48. Vandalism 49. Tran sportation 50. J u st for Fun 5 1 . Good Service 5 2 . B oth H a n d s 5 3 . L o n g Odds 54. The Talkative Guest 5 5 . Uranium Triangles 56. Men on the Moon 57. A Fri e ndly Teller 5 8 . Many a Mickle 59. A Tale of Woe 60. Quite a Family 6 1 . Numbers, N u mbers 6 2 . The Ye ars T h at Count 63. A Funny Fraction 64. Two M agic Squ are s

Problem Page 8 9 9 9 10 10 11 11 11 12 12 13 13 13 14 14 15 15 15 16 16 16 17 17 17 18 18 18 19 19 19 20 20 20 21 21 22 22 22 22

Solution Page 48 49 49 49 50 50 51 51 52 52 53 53 54 54 55 55 56 56 56 57 57 58 59 59 59 60 60 61 61 62 62 63 63 64 64 65 65 66 66 66

65. The J oker's Wild 66. Easy Come Easy Go 6 7 . A B ug i n a B arn 6 8 . Wrong Number 69. At the Show 70. The Penalty Clause 7 1 . Consecutive Squares 7 2 . Think o f a Number 73. A Deal in Stamps 74. A Serial Number 75. An Evening Out 7 6 . A Matter of Age s 77. Dropping in on Le n 78. Cards on the Table 79. A Bus Ride 80. You're Sure I f in Doubt 8 1 . Cre epy Crawlies 82. Family Numbers 83. Too Many Girls 84. What? No Zobbli e s ? 8 5 . The Census 86. A Whiz Kid 87. Tickets in the Sweep 88. Progre s s 89. Three Times 90. A Touching Tale 9 1 . F u n for Some 92. He B roke the B an k 93. Do It Yourself 94. A Bug for the Birds 95. A Lucky N u m ber 96. A Matter of Time 97. The Poster 98. At the Casino 99. No Direct Road 1 00. A Nest E gg ALPHAM ETI C S APPENDICES

Problem Page 23 23 24 24 24 25 25 25 26 26 27 27 27 28 28 29 29 29 30 30 30 31 31 31 32 32 32 33 33 33 34 34 34 35 35 35

Solution Page 67 67 68 68 69 69 70 70 71 71 71 72 72 73 73 74 74 75 75 76 76 76 77 77 78 78 79 79 80 81 81 82 83 83 84 84

87

93 97

CHALLENGING MATHEMATICAL TEASERS

TEASERS 1 . PE C KING ORDER

Four sparrows fou n d a dish of seed, Fine birdie food, n o common we e d . S a i d Pi p : " I n t u r n e ach take t w o grai n s A n d t h e n a third of what remai n s . I t ' s me as first, t h e n Pep, t h e n Pop, With Pap the last. And then we stop . " B u t Pap cried out: " I t i s n 't fair. Mine's two see d s less than half Pep ' s sh are . " O l d Pip w a s boss, h i s word w a s law, So little Pap got nothing more . Poor Pap, h i s share was rather small ! How m any seeds were there in all? 2 . FOR THE ONE THAT GOT AWAY

Steve was picking out some trout fli e s from a box on the counter. " Do you have those new lures you told me about?" he aske d . "They sounded interesting." "All gone . I only got i n a few to see how they'd sell, and I 've ordere d more ," Clem replie d . "I sold the lot at as many cents e ach as the squ are of the number I h a d . " Steve smiled. " I 'll take your word for that, b u t I would like a few when you h ave the m . " " O kay, a n d y o u c a n figure o u t the price yourse lf," Clem told h i m . "All the same as before , and my percent age m arkup was what I had to pay in cents per lure . " What w a s h i s selling price?

2

CHALLENGING MATHEMATICAL TEASERS 3. LONG ODDS

C h arlie stopped at the table. He drew four cards from the little heap there , and turned them up. " Look, Pam ! " h e excl aime d . " T h e four ace s . " H i s wife smil ed. " You don 't h ave a complete d e c k there . T h e ki ds were playing a n d some cards are missing," she told him. " B ut you wouldn't do that agai n in a thousand years . " " W a n t to bet?" Ch arlie had checked t h e deck a n d done some quick figuring. " I n fact the odds were exactly a thousand to one agai nst picking all the ace s from the cards that were there . " H o w m any cards were there? 4 . LIKE FATHER LIKE SON

Sam smil ed. " You 're a p u z z l e fan , so what about this?" h e aske d . "If you t ake your age from the squ are of the sum of its digits you get Ann's age . And the same goe s for my age . " "That's gre at, Dad, a n d you 're right," replied Mike , checking on a scrap of paper. " B u t you n e e d something else. The squ are of the s u m of the four digits in your age and mine is j u st one l e s s than twice the total of our two age s . " Well, h o w o l d w a s Ann? 5. WHO SPILLE D THE COFFE E ?

H arry had an i nvoice in h i s h a n d . " Look a t thi s , " h e s a i d . " Someone h a s spilled coffee on it, and n o w I can't m ake out the total amount." Susan looke d . " T h at ' s too bad, but it wasn't me," she told him. " It's for one lot of knives at $2.96 e ach , but the n u mber invoiced i s unreadabl e . " " T h e total is n o t completely obliterate d . " H arry pointed to the e ntry. " I t was for a 4-figure nu mber of dollars and some cents. Only the fi rst figure of the dollars and the last of the cents are illegible. You can still see 704 of the dollars, and a 9 as the first figure of the cents. We can figure it out." You try.

TEASERS

3

6. SO FAR APART It was a cold, wet Sunday afternoon, and of course there was nothing worth watching on TV. But Ted s e e m e d quite h appy at the old desk in the corner. Suddenly h e put down h i s pen . " I 've figured out some­ thing with our phone n u mber and Sonia's," h e said. "The girl you met i n Au stri a?" Jeff aske d . " Okay let's h ave it." Ted grinned. " It ' s only m ath. Seven times the cube of her 7-digit number i s e xactly three times the seventh power of the 4-figure part of ours." He was right, so wh at were the n u m bers? 7. ALL TWOS

Jane held up a couple of in voice s . " Look at these tot als, both for $222 . 2 2 . Three ite m s in each, and all at different prices." H arry looke d . "That's a coincidence," he comme nte d . " B ut there 's more to it t h a n t h a t . F o r e ach of the s i x i t e m s t h e dollar amount is the squ are of what's i n t h e cents colu m n ." What were the amounts? 8. THOSE SHARE S " I did agree to be E l mer's executor," s a i d Don , " b u t I never re ali z e d what an oddball the old chap was." Clare smiled. " H e was a de ar, and anyway you only h ave to see that h i s wishes are carri ed out." " Only ! " Her husband chuckled. "H ere ' s j u st one ite m in his will. He left h i s 409 Cosmos sh ares to h i s three sons, on the condition that the squ are of the number of one son's sh are s be equal to the product of the other two. And I can't figure it out." There was nothing wrong with E lmer's arithmetic, so how would those sh ares be apportioned? 9. YOU HAVE TO KNOW HOW

"Try m ultiplying your age by seven, Dad," said Len. "The quick way."

4

CHALLENGING MATHEMATICAL TEASERS

Tom smil ed. "There ' s only one way I know, but I ' m not sure I can do it in my h e a d . " " Sure y o u can," Len told h i m . "Just t a k e t h e digits i n the opposite order and p u t the proper digit between the m . " T h e b o y w a s right . So h o w old w a s T o m , a n d wh at was t h at special digit? 10. SAVINGS

" So you let Stan e m pty his money box. I gue s s he'd lost i nterest anyway," said Sally. "Was there much in it?" " Not m u c h , but it's fu nny the way it was ," her husband repli e d . "The dimes, nickels and qu arters were all prime n u m bers, all different. And there were as many pennies as the n u m be r of dimes m u ltiplie d by the total of dimes and nicke l s . " "What about the qu arters?" S a l l y asked. M i ke smiled. "A good questi o n , but figure it out for yourse lf. There were twenty-four more pennies than q u arters, and that's all the money there was." How much did the box contain? 1 1 . A RANDOM MAI LING

"A little j ob for you," said Sam, h a n d i n g George a c ard i n d e x box. " Pick at random and mail our new folders to a fair s a mpling of these people . About five perce nt, say." B u t George h a d a plan. He would pick the first, miss o n e , pick the next, m i s s two, pick the next, miss thre e , and so o n . A n d h i s sch e m e worke d , for his fi nal card was actu ally the last card of the index. Furthermore, when h e checked h e fou n d he had picked exactly fi ve perce nt of the total n u m ber. How m a n y had he picked? 12. NO COM PUTER FOR TH IS

Ron had been busy at the table quite awh ile. "I give u p ! " h e exclai med suddenly, throwing down his pen. " You'd need a computer."

TEASERS

5

"Come on, it can't be th at tough , " h i s father told h i m . " B ut wh at' s t h e proble m?" " Something our math teacher gave u s , " the boy repli e d . " W e h ave to fi nd an 8-figure n u m b e r that is one-third what you get when you add the squ are of its first four digits to the squ are of the last fou r . " Certainly a tough probl e m , but it c o u l d be s o l v e d with­ out too much n u m e rical working. Try it ! 1 3 . AT THE D INER

The place was crowded, and they were lucky to get a table to themselves. Now they sat there impatiently, six hungry young people , waiting for their food. Ron sat on the left of the girl who sat on the left of the m an who sat on Joan's left, and Ann sat on the left of the man who sat on the left of the girl who sat on the left of the man who sat on the left of the girl who sat on the left of Pam 's husband, while Steve sat on the right of the girl who sat on Harry's right. Pam was not next to her husband, but which of the three men was he? 1 4 . THEY DON'T COME SINGLY

Fred put down h i s pe n . " N o wonder my total was $ 1 3.33 wrong," he said. "I entered the dollars as cents and the cents as dollars for Bradley's check." Helen smiled. " T h at wasn't very s m art . B u t I though t t h e y p a i d q u i t e a b i t more . " " You're right," Fred agree d . " B u t it must h ave been my off day. I entered the same payment twice, e ach time m aking the same fool mi stake . " What was the actual amount? 1 5. WHO GLUB S GLYGONS?

The four fallians had been gathering glygo n s for quite a while when Flab called a halt. " We've got ne arly a thou sand," he declare d . " L et's sh are them out." As the others watch e d , Flab counted out a qu arter of the total for h i m s e lf. " Here 's one e ach for you to glu b , " he told the m , flipping out three m o r e from the h e a p . " Now it's Flib's turn . "

6

CHALLENGING MATHEMATICAL TEASERS

Flib took a qu arter of what re mained, and also one e ach for Flob and Flub. Then Flob took a qu arter of what Flib h ad left, and one e xtra glygon for Flub. And fin ally Flub took h i s qu arter. "That's fi n e , " said Flab, one tentacle re aching into the much-depleted heap. "We can divide the re st equally among u s . " H o w m a n y h a d t h e y gathered? 1 6. THRE E IN A ROW "That's interesting ! " Ray leaned back in h i s ch air. " I f you write down t h e age s o f o u r three kids in a row and divide that 3-digit n u m ber by the sum of their age s you get thirty-two . " Fran shook h e r h e a d . " You and your figure s . What ' s so special about th at?" " Well, it's your age , for one thing," her hu sband replied. " M aybe you forgot . " H o w o l d were the children? 17. A SWITCH M arti n had a piece of paper in his hand. " H ere ' s some­ thing I j u st figured out about our new phone n u m ber," h e said. "The complete seven figure s . " " I h ate it. It was so m u c h e asier to re member a word or even letters , " S ally re pli e d . " B ut what did you fi nd?" " Look," M artin told her, pointing to the paper. " I f you switch the two part s , putting the four digits in front, you get one more than twice our complete n u mber." So what was it? 18. JUST TRIANGLE S Mike put down h i s ruler. "That's it. Look at the tri an­ gle s I 've drawn , Dad," he said. "Each are a is exactly two­ thirds its perimeter." " You mean i n squ are inches and i nche s?" Victor looked. " O n e of them see m s ki n d of t h i n . " " Sure , b u t it is a tri angle. A l l the sides of t h e s e trian­ gle s are whole n u mbers of inches." What were the dimensions of Mike 's tri angles?

TEASERS

7

1 9 . AN ANCIENT PROB LE M

Peter put down h i s pe n . " It ' s j u st something I 've fig­ ured out on age s , " he said. " M y age and the age s of Aunt B e ssie and Uncle Joe . When you total any pair of those age s you get a pe rfect s q u are . " H arry stu died t h e sheet o f paper on t h e table. " A n odd coincidence," he told the boy. " B ut there ' s another one. Those three age s also total the squ are of your cousin Sally's age , and she's not yet in her t e e n s . " W h a t were the four age s? 20. S I M I LAR B UT D I F F E RE NT

Doug turned wh e n h i s father c a m e into the roo m . " Look at the two tri angle s I 've drawn ," h e said. "The sides are all exact nu mbers of inches. " " U sing m y b e s t drawing board, e h ?" T o m smiled. " B ut what's so special about them? I can see they're ex actly similar, one bigge r than the oth er." "That's only part of it, Dad," the boy repli e d . " E xactly the same shape, but two si des of one are the same l e ngths as two sides of the other. And there ' s nineteen inches difference betwee n the sides that are not the s a m e . " What were the di m e n si o n s? 2 1 . JUST FOR K I D S

S t a n put d o w n the b i g carton and flopped into a ch air. "I left one out in the car," he said. "All cra z y hats, and I hope they sell . " " Sure t h e y will , " h i s wife replied. " T h e ki d s a s k for them all the ti m e , the crazier the better. I hope you got a fair selectio n . " " A s y o u said . " Stan nodded. "Three style s, three diffe r­ ent price s. All in even dozens, and I bought as m a n y at each price as t h at price in cents . They averaged ex actly h alf a dollar e ach for the lot . " How m a n y h ad h e bought? 22. THE PATIO

B i l l was busy i n his backyard when Fred arrived. " I j u st fi nished our new patio," h e s a i d . " Now i t only n e e d s s o m e flowe rs i n t h e middle . "

8

CHALLENGING MATHEMATICAL TEASERS

I t was a most u n u s u al patio, but then Bill was always eccentric. He had paved three squ are are as, with corners meeting so th at they formed the sides of an unpaved, enclosed tri angl e . " Fi n e and d andy, but it does look a bit odd," Fred commented. " R e ally three little patios . " "That's right," Bill agree d . " I m a d e t h e i r are as 1 96, 9 7 a n d 4 1 s q u are feet." His friend smiled. "What about the flower bed in the middle?" A good question ! What was its area? 23. SQUARES AND SQUARE S

Sam was busy at h i s desk whe n Joe went i n , crumpled paper littering the rug around h i s ch air. " Tryi ng to do J ack's homework?" Joe asked. " Kids are s m art these d ays . " " N 0, i t ' s something I m ade up myself, a n d i t ' s a dandy," Sam replied. "If you add our house n u mber to the product of t h at n u m be r and Jack's age you get twice the squ are of my age . But if you add Jack's age to that same product you get twice the squ are of m y wife ' s age ." It was a n e at little problem. Wh at were the age s of Sam and h i s wife? 2 4 . WRONG B UT RIGHT

Wendy watched with growing i m patience as the clerk m ade out the check for her purchases. " S ay ! " she ex­ claimed at last. " You multiplied the three amounts in­ stead of adding t h e m . That's crazy." " Sure I did, but it's okay," the young man repli ed. " The total comes to $ 5 . 70 either way. Add the m u p yourse lf. " I n d e e d he was righ t ! So what we re the individ u al amounts? 2 5 . THE STAMP

" S o that's a Kalotan stamp," H arry commented. "A gre at design . Do you h ave any m ore?" " Not now. I did h ave a 3 kuk, a 9 kuk, a 10 kuk, a 1 2 kuk, a 1 3 k u k and a 1 4 ku k," A n d y replied. " B ut at t h e airport I u s e d five of t h e m t o m a i l a couple of postcards.

TE ASERS

9

One took twice as much postage as the other, and that left me only this one sta m p . " What w a s its value? 26. A TALE O F TWO GUYS

M ary was still busy with the d i s h e s when Fred c a m e i n . " I guess the power's been off, " he c o m m e n t e d . " N o dish­ washer." " Most of the day," re plied his wife . " I t only c a m e on j u st now, but how did you know?" "They were cleari n g up down the road around a new pole and I talked to the l i n e s m e n there , " Fred told her. " I t see ms they fi xed two guy wire s o n one s i d e , both fro m t h e same band 28 fe et up the pole , with t h e i r e n d s 1 50 feet ap art at the grou n d . Their two le ngt h s total j u st 250 feet, and I could see they were both taut." Assu m i n g the ground was level, what would the lengths of the respective guy wires be? 27. JUST JUN K Sally flopped into her favorite ch air. " I 've j u st cle ared all the j u n k in the attic," she said. " Wh at a j o b ! There was an insurance policy, but I gu e s s it's fi n i s h e d . " " Probably on the o l d house, the one I kept b e c a u s e o f i t s fu nny nu mber," Greg told her. " B ut y o u got r i d o f t h o s e fi re h a z ards up there , so it wou l d n 't m atter so much anyway . " H i s wife smil ed. " W h at was peculiar about t h e n u m ­ be r?" "Well, I though t it od d , " Greg replied. " I t had seven figure s . The fi rst, third , fifth , and last all one s . W h e n I tried to divide it by seven, and by e l e v e n , and by thirte e n , I w a s o n e sh ort e ach t i m e . B u t I forget w h a t t h e other three figure s were . " You fi gure t h e m out. 28. A SIM PLE ROUTINE Peter put down h i s pen. " D a d , " he said . " Did you know Uncle Fred's car plate n u m ber i s like yours?" " H ow's th at?" John asked. " I 've got a 4-figure n u m ber, but he has five fi gure s . "

10

CHALLENGING MATHEMATICAL TEASERS

The boy gri n n e d . " I mean in one way," he explained. " You've got 6078. I f you squ are its last three figures and then su btract the other p art, you get the origi n al n u mber agai n . " H i s father w a s checking. "Th at's right," he declared. " T h e squ are of the last three i n 6084, and su btracting 6 you get 6078. So what?" " Well, the same routine works with Uncle Fred's num­ ber." He was right on that too, so what was his uncle's num­ ber? 29. CO M M E MORATION STAMPS

She had locked the door behind the last customer, and now Gwen was h e lping her husband tidy up their little store . " You know the packets of comme moration stamps we m ade up?" s h e asked. "We 've got about fifty packets left . They don't s e l l . " " J u s t t h r e e different selectio n s . T h r e e 4¢ , t w o 5¢ and fi ve 1 0 ¢ ; seve n 4¢, three 5 ¢ and three 1 0 ¢ ; one 4¢, five 5¢ and four 1 0 ¢ . " Paul chuckl e d . "I still re m e m ber th at chore . Any sugge stions?" " Package those re m aining stamps agai n but in sets of three stamps e ac h , one stamp of e ach denomination," Gwe n replie d . "Th ere were no spare s left over, so let's check wh at we've got . " It d i d n 't take long to do t h a t , and then Paul did some figuring. "I gu e s s you had it figured out anyway ," h e told her. "We 're j u st right, not too few and not too many for your sch e m e . " H o w m a n y sta m p s did t h e y h ave left? 30. A LMOST NE IGHBORS

John put down the letter. " From an Elsie Brandon. She lives on your road, n u m ber 74 , " he said. " M aybe you know her." "I don't recall the n a m e , but it' s a long roa d . " Ke n shook his h e a d . " B ut that's fu nny. We h ave no n u m bers m i s sing on the street, so there are exactly as many below her n u mber as there are above ours . " "Then there ' s another coincidence." John w a s doing some figuring on a scrap of paper. "All the n u m bers above

TEASERS

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yours, added t o all below hers, total j u st h alf the total for the whole road . " What w a s Ke n ' s n u m ber? 3 1 . RE A L E STATE

" I bought that piece of land I told you about," said George . "An exact nu mber of squ are yard s , not far s hort of twelve acre s . A snip for developme nt." " Let's hope you 're right," Jack commented. " It did sound an odd shape . A sort of isosceles tri angl e , you told me." His friend nodded. "Well, almost. I t ' s a perfect tri angl e , b u t there are s e v e n yards betwee n the sh ort and middle sides, and seven between the middle and the long. And it j ust h appe n s that the sides are all exact n u m bers of yards." Located where two highways meet, that land could h ave possibilitie s . What would its dimensions be? (Note : There are 4840 squ are yards in an acre .) 3 2 . GRE ETINGS

Pam put down her book. " I 've j u st reali z e d it's Uncle Tom's birth day today," she sai d . "We must send the old chap a telegra m . " " Okay," Doug agre e d . " I 'll make it o u t a n d you can phone it." Those greeti ngs are always difficult to word , and Doug was scribbling for quite a whi l e . Then he looked up with a smile . "That's fu nny. The squ are of Uncle Tom's age is j u st one less than the difference between the cubes of Mark' s and Judy's age s . " " B ut there 's only o n e year betwee n the m , " h i s wife obj ected . "How come?" "The cubes of their age s , " Doug replied. " You check it." How old was Uncle Tom? 33. AWAY FROM IT ALL

" We 're out i n the sticks, but we do h ave our pick of four h andy village s whe n the rustic scene palls," said Mike . " M aybe you s a w t h e two roads that branch off to the northward j ust opposite our gate s . "

12

CHALLENGING MATHEMATICAL TEASERS

" Ye s , I d i d , " George replie d . "Of course I came through Brill on the road that goe s on to Sefton." M i ke nodded. " Th e other two roads go to Carlow and Lawanda, and all three are dead straight. We h appen to be exactly the s a m e distance fro m all four village s . " " S o t h o s e are y o u r cultural cente rs!" His friend smiled. "I didn't think much of Brill." "Well, we h ave t h e choice . As the crow flie s , Carlow is thirty ki lometers fro m both Sefton and Lawanda, and Lawanda fourtee n kilometers fro m Brill." How far was e ach from Mike ' s gate s? 34. DOUBLE THE ODDS

J i m held out a little cloth b ag. "Just m arble s, Dad. About forty of the m , some re d , the rest green," he said. "No peeking, but take one . " D o n dipped i nto t h e b a g and pulled o u t a gre e n . Trying again h e got a second gre e n , which he placed on the t able beside the fi rst. " You can't h ave many re ds," he com­ mente d . " Wh at are the odds agai nst a third gree n if I try once more?" The boy thought a moment. "Just twice what they were agai nst a re d for your fi rst draw, " he repli e d . H ow m a n y m arble s of e ach color d i d J i m h a v e there? 3 5 . BOUNTY H UNTERS

" Your garden was full of kids yesterday," said John. " A birthday party?" " Sort of. Re ally a gru b hunt combined with a party for Doug." Bob chuckle d . "There were fifty kids in all. " " Sm art i d e a , " John commented. " A n y pri ze s?" " Sure . A qu arter for every caterpillar kille d. It's odd that the boys all killed the same n u mber, and so did the girl s , but not the s a m e n u mber as the boys ," Bob re plied. " B ut e ach boy h ad to pay me back one penny per grub for all the caterpillars the girl s ki lled, and e ach girl the same for all that the boys ki lled. It cost m e only nineteen cents per boy and eleven cents per girl , and they killed around a th ousand in all . " H o w m any gi rls were there?

TEASERS

13

36. THE M URAL

Steve stopped in the studio doorway. " So t h at's t h e gre at m ural , " he said. "An abstract across the complete wall. It's certainly got something, but what the heck does it me an?" " More than you may think," Ted told h i m . "Tell m e what y o u see . " H i s friend thought a moment. "Just lots of rectangle s , " he repli e d . "They a l l s e e m t o be grouped i n pairs . " T e d smiled. " Th at's n o t a l l . A l l t h e s i d e s are whole n u m ­ bers of i n c h e s , and in e very p a i r the are a in square inches of e ach equals the peri meter of the other rectangle i n the pair. 1 drew all the pairs for which that can apply. " What were t h e d i m e n sions o f a l l h i s rectangles? 3 7 . A WOMAN'S JOB

" I 'll fi nish the m . " Ann went over to where her h u sband worked we arily, with envelopes and flyers all around h i m . " O kay." Bill straighte ned up thankfu lly. " I figure I 've been on the j ob alone for j u st fi ve-twe lfth s as long as you 'd h ave taken to do the lot, so m aybe it's time you did some . " A n n s m i l e d . " We 've d o n e t h e s e often enough before , so you should know," she said. "Anyway it's a wo m a n ' s j ob . " For quite a w h i l e there w a s o n l y the s o u n d of paper sliding on paper in the little office. And then Ann stuffed the last flyer into its container. " I 've been figuring too," she declare d . "We would h ave saved j u st fifty-two min­ utes if we had worke d together right through . " B i l l p u t down h i s m agaz i n e , gl anced a t h i s watch , a n d scribbled on a scratch p a d . "You're right," h e told her. "And 1 would have done exactly h alf what you've actu ally done j u st now." Next time they m ay h ave hired help. But how long h ad the complete j o b take n them? 38. EACH TO EACH

"Well, John, it's good to see you agai n . " Bob greeted h i s brother. " I forgot to write thanking y o u a l l for your card s . "

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CHALLENGING MATHEMATICAL TEASERS

"Me too . " John smiled. " B ut, talking of card s, the Doo­ little family will re ally be boosting the card indu stry if we conti n u e the way we 're going. This X m a s I figure there were eighty-four more cards exch ange d among all of us than the previous X m a s , and there were eighty-four more th at X m a s than the one before . " Bob chuckle d . " S o we buy s h a r e s . It's a n i c e idea, though, even with all the babies i ncluded i n the routi ne, th at e ach m e m ber of the combined families sends a card to each othe r me m ber." How many cards had been exch ange d th at last Xm as?

39. THE JEWE L BOX

Jim put down the letter. "A rush order, and clear on all details except the most i m portant one," he said. " H e w a n t s one of t h o s e squ are sn ake s ki n j ewel boxe s , the one h e says h e saw here last mont h . " "They're all different s i z e s , and we're supposed to re­ member." Paul chuckle d . " Let's ask for m ore informa­ tion . " " H e ' s traveling, so w e can't contact h i m . But he wants to pick it up next week," decl ared his p artner. "We h ave to set an initial i n diamonds on the lid before then. Its center must be five c m fro m one corner of the lid, thirteen cm fro m the opposite corner, and eight cm from one of the corners betwe e n . " P a u l thought. " Not to worry , t h e n . There ' s o n l y o n e of e ach s i z e , and w e can fi gure out what the size must b e . " H e was right, so what would be the are a of t h e top?

40. PETER'S PE NNIES

Peter had all his pennies laid out compactly on the table . " Look, Dad," he said. "They m ake a pe rfect square . " Ch arlie smil ed. " T h at ' s right," he told the boy, bre aking up the squ are and moving the coi n s i nto a different ar­ range ment. " You 've got j u st e nough to m ake something even better. There you are , a pe rfect hexagon like the cell i n a h o n e yco m b . "

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"That's gre at . " Peter w a s impre s s e d . " Let's fi nd some more pennie s , for a bigge r squ are that will m ake a bigger hex thing." "I gu ess you'd need an awfu l lot of penni e s . " Ch arlie shook his h e a d . " M aybe I can fi gure it out." How many penn ies would be needed to carry out the same routine on the next bigger scale , the coins all flat on the table and packed as before? 4 1 . SO VE RY SI M PLE

Ch arlie handed back the sheet of paper. "Come on, that's an e n ormous n u m ber," he declare d . " You can't ex­ pect me to multiply it by twenty-n ine in my head." " Why not? It's e asy if you know how." Mike chuckled. " You only have to write the same extra digit at e ach end. And it's the s m allest number th at works th at way for twe nty-nine . " What was Mike ' s n u m ber? 42. POWE R PLAY

Harry h ad been quiet for quite a while , busy with h i s homework. " Look, Dad," he s a i d suddenly, holding up a sheet of paper. " I 've figu red out something about those three n u mbers . " Bruce looke d . "That's our own house nu mber, a n d your Aunt Jane's, and U ncle Tom ' s , " he commented. " B ut what's all the figuri ng?" " It connects the m , " re plied the boy. "The squ are of ours i s twice the cube of Aunt Jane's n u m ber, and it's three times the fifth power of Uncle T o m ' s . " It w a s h ard ly regu lar homework, b u t the b o y was right. Wh at were those n u m bers? 43. THE COLLE CTOR

Walt was busy with his stamps when Jill looked i n at his place . "I saw you coming out of Bretts last week," she said. " Di d you fi nd any b argains?" " N othing very special, but I did buy quite a lot-stamps I needed to com plete certain sets , " her brother repli e d .

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CHALLENGING MATHEMATICAL TEASERS

"Just one h u n dre d in all, and at four price s : 59¢ , $ 1 .99, $ 2 . 8 7 and $3.44 e ach . " "A l o t o f m o n e y , " J i l l commented. " How much?" Walt smiled. " One hun dred bucks exactly, and I guess I couldn't really afford i t . " H o w m a n y had h e bought a t e ach price? 44. A MATTE R O F SQUARE S

Ch arlie stopped outside the entrance . " Anyway it does h ave an interesting n u m ber," h e commented. "Quite u n i q u e , i n fact." " Looks very ordi n ary to m e . " Greg shook h i s head. " B ut you know how I am about figure s . " " Deplorable i n a man o f your age ! " Ch arlie smiled. " B ut I 'll tell you . If you total the squares of its three digits you'll get exactly h alf that nu mber." What was the n u mber? 45 . THE ONLY TRE E

Ge orge took out h i s little book. " Let's h ave the nu mber of your new h o u s e , " h e said. "I drove through your road yesterday, but I couldn't re me mber the n u m ber." Steve smiled. "I gu e s s they all look much the same, but we've got the only big tree in the development . " " I noticed i t , " George told h i m . " So you're o n t h e north side." " T h at ' s right, all even nu mbers a r e on o u r side ," Steve repli e d . " B u t you figure it out yourse lf. All the even n u m ­ b e r s above o u r s t o t a l h alf the s u m of the e v e n nu mbers below ours, and we've got a 2-figure n u m ber." What was it, then? 46. IT M A K E S SENSE

If one two one plus two one two Is three t i m e s forty-three , T h e n w h a t can one one one plus one O n e one plu s one o n e be?

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4 7 . GRAN DAD'S B I RTHDAY

" Don't be l ate now," said S u s a n . " Your grandfather's coming over. It's his birthday, you know." Doug nodded. " Sure, and I m ade up something on his age . I f you add up all the age s I 've been, including my age now, you get one year m ore than h i s age , " h e re plied. "And the total of the two figures of h i s age i s m y age . " What were t h e two age s? 48. VAN DALISM

" I saw th at some crazy kook broke into Jake ' s pet shop and ope ned some of the cage s , " said M arth a . "He had j u st 300 birds there and more than 1 0 0 escape d . " "Too b a d , t h e y 'll mostly d i e o u t on their own , " Fred replie d , checking i n his newspaper. "They say a third of what re mained were fi nch e s , a qu arter budgi e s , a fifth canari e s , a seve n th mynah bird s , and a ninth were par­ rots. And the original n u m ber of can ari e s was three times the nu mber of p arrots that re mained . " Stupid vandali s m ! B u t o n e o f those fractions was wrong. How many canarie s e scaped? 49. TRANSPORTATION

Sam shook his head. " I t's quite a stretch , twe nty-four miles to TulIa. But I can only take ten of you , all the truck will carry . " Greg th ought a m o m e nt. "That's o k a y , w e ' l l re lay," h e declare d . "There are thirty of u s , a n d it's e s sential we a n arri ve a t t h e same time and a s s o o n as possible . I could figure out the way if I knew what speed you'll drive . " T h e old m a n chuckle d . " S h e ' s old a n d slow. I t ' s a steady drop all the way there , so we 'll do forty miles per hour. B ut coming back, uphill on that road, we'll only m a ke thirty miles per hour." " F i n e . " Greg h ad been j otti ng down some figu re s . " Some o f us can start on foot a n d be picked up along the way. We walk at four miles per hour, so we'll all walk and ri d e . " Assu ming truc k and walkers a l l started at the s a m e ti me, h ow l o n g wou ld it take the p arty to r e a c h TulIa?

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CHALLENGING MATHEMATICAL TEASERS

50. JUST FOR FUN

Walt was scribbling on a scratch pad. " R e m e mber that teaser about a phone n u mber?" he aske d . "Well, I 've m ade one about our new nu mber." "I re m e m ber. Something about the 3-digit exchange and the old 4-digit part, but I forget the details," his partner repli e d . " W h at have you figured out?" " Look. " Walt pointe d . " If you su btract the exchange p art from h alf the squ are of the other bit you get the complete 7-digit n u m ber agai n . " Wh at w a s it? 5 1 . GOOD SERV I C E

"We h ave a fine com m uter service here , " s a i d Steve . " Trains to your station all day long, and by two different route s . " " So u n d s goo d , " Fred commented. "What t i m e s do they ru n?" "The first train via Poole leaves at 6 : 00 a . m . , followed throughout the day by train s at regular intervals. Going via Tulla, the fi rst is at 6 : 05 a . m . , also followed throughout the day by tra i n s at regular i ntervals," Steve replied. "They conti nue like that right through until soon after midnight on both route s . But it's odd that departures coincide only twice, the first time when the nin eteenth Poole train and the eighteenth Tulla le ave h ere simulta­ neously." The " regular i nterval s " we re exact n u m bers of min­ ute s , so what were the two ti mes of simultaneous depar­ ture? 52. B OTH HANDS

" I ' m bore d . Give m e something to do," said Joe . "And what's the ti m e anyway?" Ben gl anced at his watch and did some quick figuring. " When I checked m y watch this morning," he replied, "the hour han d was ex actly where the min ute h and is now, and the mi nute hand was one division before where the hour h and i s now." " You m e an m i n ute division?" Joe asked. "That's right," B e n told h i m . "And both h an d s are ex­ actly at min ute divisions now . " W h a t w a s the t i m e ?

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53. LONG ODDS

Pat picked up part of the deck of cards , and examined them. " Okay , " he said. " Don't look, but draw three cards from these in m y h a n d . " J a c k complied, turning e a c h card face up as he drew it. "All spade s," he com mented. " Th at's fu nny." " I t sure i s , " Pat agre e d . "In fact the odds were ex actly 1 1 to 2 against you getting three spade s . " How m a n y cards h ad P a t picked up from th e deck, a n d how m a n y had b e e n spades? 54. THE TALKATIVE GUE ST

" Wh at a blabbermouth ! " I an exclai m e d , watching as the car turned out of the driveway. " Di d you get h i s story?" "About four cli e n t s of h i s : a banker, a curate , an actor and a denti s t . " Sally smiled. "He called t h e m Doug, Clem, Andy and B e n . But I never got which was which . " " I d i d bette r," her husband declare d . " Doug's t h e den­ tist if Ben i s the banker, but he's the actor if Andy's the curate. Ben is not the actor unless Doug's the dentist. B e n ' s the banker if Clem's the actor. Then Doug is the curate unless either Cle m i s the dentist or B e n ' s the banker, and anyway the curate is not B e n . " I an w a s quite right ! Y o u figure it out. 55. URANIUM TRIANGLES

Tom's old gardener greeted m e as I walke d d own the long driveway toward the house. "What's th at you've m ade?" I asked him. " Looks like two new flower beds, but they're big. " "That's right," Sam repli e d . " It ' s something he got from a book. He says they're uranium tri angles, with all sides exact numbers of feet and both are as exact squ are fe et." " You mean Heron i a n . " I smiled. "An old guy who lived about two thousand years ago. He was quite a m athe­ m atici an. B ut what are the exact dimen sions of your tri­ angles?" "Well, their are as are the s a m e . " The old chap gri n n e d . " E ach of them has one side sevente e n feet and another side twe nty-e ight feet. But the ir third sides are different, so m aybe you can figu re them out." You try.

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CHALLENGING MATHEMATICAL TEASERS

56. MEN ON THE MOON " What's a moon crawler, Dad?" asked Elsie, looking up from her maga z i n e . "A special sort of truck with huge tire s , " Steve t o l d her. "They're being use d on the moon . Why do you ask?" " I t ' s mentioned here , " the girl replie d . "They're going to cross the Mare Hunticus. I gu e s s that's all sand, and it's j ust 528 m i l e s wide . " "Well, that's okay for a crawler if it carri e s enough fu e l , " Steve co m m ented. " B ut I don't i m agine it can carry very much . " E lsie smiled. "That's why I aske d , " s h e s a i d . " I t says a moon crawler carri e s e nough fuel for only 3 1 5 miles i n a l l , so I d o n ' t see how it c o u l d do the trip." " W h at about setting up its own fuel cach e s e n route?" Her father was thinking. " Th at's it, so let's figure out the m i n i m u m total mile age they would h ave to do to m ake the crossing. You try ! 5 7 . A FRI E N D LY TE LLER Susan e mptied her purse onto the table . " It ' s cra z y ! " s h e excl aimed. " I h a d only $ 1 . 2 5 before w e cashed m y refu n d check a t t h e bank. Now I 've got exactly double the amount of the check." Steve checke d . " You 're in luck. But don't blame me. We've been nowhere else . " " It m u s t h ave b e e n t h e teller," S u s a n told h i m . " Let's go back." " Okay , " her husband repli e d . "I know what she did. She gave you dollars for cents and cents for dollars, and she also reversed the order of the digits for e ach . " He was right, so wh at was the amount of the check? 58. MANY A M I C KLE " Some people m ake good beggars, " said B ob, scanning the listed totals of the donations collected by his little band of helpers. E ach ye ar he and a few close friends devote much time to getting what they can from their acquaintances for the eleven ch aritie s that they support. " Good?" Alan chuckled, pointing to the list. " Look at

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those fi rst two tota l s : 24¢ and $ 1 . 2 2 . I 'd call th at beggarly beggary ! " Bob smiled. " M y brother made out the list, and those were wh at h i s two ki d s collecte d . H e put the m in sepa­ rately j u st for fun . But after the $ 1 . 2 2 e ntry every total is exactly double the previous one on the list, and that's an amazing coincide nce . " " I t sure i s , " h i s fri e n d agre e d . "I s e e y o u deduct only $52.00 altogether for expe n s e s , and then you can divide the balance exactly equ ally among your ch ariti e s . " H o w m u c h would e ach charity re ceive? 59. A TALE O F WOE

Walter shook h i s head. "I closed down last week, and I ' m broke , " he said. "The last d a y we took i n precisely $5.33, and that was the e n d . " " I ' m sorry . " Andy was shocke d . " You started w e l l , with good sales the fi rst day and people standing in li n e . " " M aybe too we ll. E very d a y after that we took in l e s s , " Walter t o l d h i m . " It was u n c a n n y , like a curse . The second day we took i n j u st one cent more than two-thirds of the first day's take , the third day two cents m ore than two­ thirds of the second day, the fourth day three cents m ore than two-thirds of the third day, and so on. E ach day another penny m ore than two-third s of the previous d a y . " H e had a r e a l proble m , and so w i l l y o u if y o u want to solve thi s ! How much had Walter taken the fi rst day? 60. QUITE A FAM I LY

" It's fu nny about B e n ' s five ki d s , " said Joh n , blessing the i n ane com m erci als that do provide some respite for conversation. " Not that we'd want five . " " I certainly wouldn't." A n n laugh e d . " B ut what's s o o d d about the m ?" " Something I noticed today about their age s , " replied her husband. "They're all differe n t . The three boys are spaced two ye ars apart, and the gi rls are too. And the square s of the boys' age s total exactly the differe nce be­ tween the cubes of the girls' age s . " A n n shook h e r h e a d , turning back t o t h e T V . T h a t at least d e m anded no thinking! What were the five age s?

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CHALLENGING MATHEMATICAL TEASERS

6 1 . N U M B E RS, N U M B E RS

Stan put down h i s p e n . " I found something fu n n y about your license n u mber," he sai d. " It's one more than what you get if you add together the square s of its two h alve s . L o o k , Dad." John looke d . " T h at ' s the fi rst three figure s and the last thre e , eh? 403 and 4 9 1 . Do you get 403490 if you total their squ are s?" " Sure , " replied the boy. " B u t d'you think there are any other 6-figure n u m bers t h at work that way?" See you m any you can fi nd ! 62. THE YEARS THAT COUNT

Sam stood the photo up again on his desk. " Ye s , that's the whole fa mily. j u st the four of u s . " "Very n i c e too . " Peter nodded. " I m e t your younger boy when he came i n today with your wife . He told me he's nine, but I gue s s your othe r son i s quite a bit older." " T h at ' s right," Sam agre e d . " I t's odd about all our ages. I f you total the squares of the boys' age s and mine, you get my wife ' s age m u ltiplied by the total of my age and their two age s . All in complete ye ars , of course . " Peter saw what he m e ant, b u t never managed t o fi gure out those age s . W h at were t h e y? 63. A F UNNY FRACTION

Mike pointed to an i nvoice th at h ad come i n the mail. " T h at ' s an odd amount," he sai d . " I f you shift the cents and put th e m before the doll ars you get j u st fi ve-eighth s of it in cent s . " Ch arlie looke d , a n d did some fi guring on the b a c k of an e nvelope . "That's right," he declare d . " I guess it's the s m alle st amount that works th at way . " H e w a s right t o o , so wh at w a s the amou nt? 64. TWO MAGIC SQUARES

We h ave two regular M agic Square s . E ach entails a chec kerboard-style array of po sitive con­ secutive n u mbers i n squ are form ati o n . E ach row, e ach

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colu m n , and e ac h m a i n di agonal m u st add up to the same m agic total . The m agi c totals are the same , and i n e ach squ are 54 is the gre atest n u m ber. Yet the M agic Squ are s are not the same ! What must their re spective s m allest n u m bers be?

65. THE JO KER'S W I L D

Jack riffled t h e card s expertl y . " Let's t r y something different tod ay," h e said. " Okay . " Bob smiled. "What's the ide a?" " Si mple," h i s fri end re plied. "At the e n d of each game the loser pays exactly a third of the cash he's got t h e n . " It sounded fair e nough , so the t w o started playing. But after a while Jack pushed back h i s chair. "That's i t , " h e declare d . "You've now got precisely t h r e e times w h a t I h ave, and I 've lost almost e xactly four bucks. And you won each of the last few game s . " "Just l u c k . You w o n every g a m e before t h o s e , " Bob re minded h i m . "In fact we 've both won the same n u mber of games." How much did e ach h ave when they starte d?

66. EASY COME E ASY GO

This was one of the t i m e s B i l l fou n d it h ard to ke ep his temper i n t h e face of h i s p artner's peculi ariti e s . B ut most of the money was B e n ' s , a n d things did seem to work out right i n spite of everyt hi ng. " So you ord er a lot of socks at $ 1 .3 5 a pair," he com­ mented. " M aybe fi v e h u n dred dollars worth for all I kn ow, but you h ave no idea how much exactly or how m any." B e n smiled. "Th at's righ t, except I d i d n 't spend that much," he sai d , sprawled in h i s chair quite at ease in that rather absurd situati o n . " You worry too much about de­ tails, and there'll be an i n voice . But I did n otice when the gu y wrote it down that the nu mber of dollars in the total was th ree more t h a n the squ are of the n u m ber of cents. That was with out any tax or discount." How many pairs h ad h e ordered?

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CHALLENGING MATHEMATICAL TEASERS

67. A BUG IN A B ARN

When Tom looked i nto h i s long barn he found Bob busy with a m e asuring tape . "What's going on?" he asked. " Checking wh at our teacher sai d , " replied the boy. " Say a roach ran from that door to the other one, touching the front and back walls on the way, how far would his short­ est possible route be?" Tom smil ed. " B eyond m e , " he said. "I guess it m u st depend on the di m e n s i o n s . " "That's right . " Bob glanced at the b i t of p a p e r in h i s h a n d . "The barn i s thirty-two fe et, front to b a c k , and ninety feet long. That door at this end i s thirteen feet fro m the back, and the door at the other end is eleven feet fro m the front, t aki ng the middle of e ach . " What do y o u say? 68. WRONG N U M B E R

" I 'll h ave to g e t m y n u m ber change d , " said Joe. " I 'm called at all hours, but it's never my number they want." "That's tough . " M i ke nodded sym path etically. " Must be some n u m ber very like yours . " " T o o darn e d li ke , although actu ally j u st three ti mes m y n u m ber. I fou n d o u t i t ' s a girl s ' residence, to do with the u niversity. Me, of all old sinners ! " Joe laugh e d . " If you shift the third figure of my n u m ber and put it fi rst, and then close up the other three figures you get the n u m ber of that hen hove l . " So what w a s his n u m b e r? 69. AT THE SHOW

Keith h a n d e d back the program . "I don't like going to a show i n such a big group, but 1 guess you got a special rate . " " Sure , b u t w e re ally were n 't t h at m a n y . If t h e whole club had gone there would h ave been getting on for a h u ndre d , " re plied A n n . "They gave u s one complete row, and we all drew for seats. M arilyn and 1 got the very end seats at opposite ends. The fi ve B arton boys had come and o n e of t h e m sat beside m e . "

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" You liked that." Her father chuckle d . " It ' s fu n n y , the odds were exactly 2 to 1 against there being a girl at e ach end. " How many were there in the group? 70. THE PENALTY C LAUSE

" You know those old rose stan dards you want dug u p , " s a i d Peter. " I 'll t e n d e r for the j ob . " " Okay, son . " H i s father laugh e d . "What's your offe r?" The boy was re ady. " I get thirty-seven cents for e ach, and I 'll do the lot i n not more than two hours . " " I 'll b u y that," declared h i s father. " B ut you'll p a y a pen alty if you exceed the two hours. There ' s ne arly n i n ety of them in all. For the fi rst minute over the time you'll pay a penny, for the second minute two cents , for the third minute three cents, and so o n . Part of a minute counts as a minute . " So Peter starte d , h i s father checking the time very care­ fully. And, when the j ob was done, Peter agree d that a net amount of $7.00 was due to h i m . H o w m a n y rose standards were there? 71. CONSE C UTIVE SQUARE S

Tony looke d up when h i s father came i n . " S e e t h i s , Dad," he said, h o l d i n g o u t a sheet of p a p e r . "That's a very special number." " Seven digits, and it e n d s with 6789," Sam c o m m e nted. But that doe s n 't m ake it special. There are lots of n u m ­ b e r s w i t h 6 7 8 9 as ending." " Sure there are ," Tony agree d . " B ut this one i s also the sum of the squares of three consecutive n u mbers. I j u st figured it out . " T h e boy w a s right, so what w a s his n u m ber? 72. TH I N K O F A N U M B E R

H arry s m i l e d as h i s wife searched her b a g angrily . " You drag me here to buy expensive paving sto n e s , and then you don't know how m a n y you n e e d . " he said. " Surely you can re m e m ber."

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CHALLENGING MATHEMATICAL TEASERS

" O f course not . " Fran shook h e r head. "That's why I wrote it down . B ut I do recall noticing the paved are a was to be ex actly seven t i m e s some other number in squ are fee t . " " Most helpful if y o u h a d n 't forgotte n t h a t key nu mber too . " H arry chu ckle d . " B u t I ' ll tell you something. The bit of lawn i s j u st eighteen feet by twenty-thre e , and the paved are a will be less than it." The com plete patio was goi ng to be squ are , with the ston e s laid out arou n d the grass. Wh at was the key num­ ber Fran had forgotten? 73. A DEAL IN STAMPS

" More stamps?" asked Conni e , fi nding her husband busy at the table . " B ut I gu e s s t h e re are worse hobbies." " I ' m gl ad you fe e l that way." Rick looked up, smiling. "I j u st blew about thirty-two bucks on these two lots . " C o n n i e p i c k e d up a s t a m p . " T h at's pretty," she said. "Any special bargain s?" " Not re ally, not at those prices, but it's fu nny how the prices came out," Rick repli e d . " For that lot I averaged as m a n y cents each as the n u mber of stamps in the lot. This s m aller lot cost me three dollars and nineteen cents l e s s , but they averaged five t i m e s as m any cents each as the n u m ber i n this lot . " H i s wife shook her h e a d . T h a t wasn't her hobby either. Maybe you can fi gure out the detai l s . 74. A SE RIAL N U M B E R

Peter put d o w n h i s p e n . " You re me mber th at o l d dollar bill you gave me, Dad?" he asked. " I t's got an intere sting serial n u mber." " S o you didn't spe nd it yet . " Tom glanced at the bill lying there on the table . " E ight digits , and I see the fi rst and last digits are the s a m e . Is that wh at you call i nter­ esting?" " Come on, Dad, I ' m not that stupid . " The boy grinned. " L ook. The complete nu mber is exactly the fifth power of wh at you get when you write those first and last digits side by si d e . " W h a t w a s t h e serial n u m ber?

TEASERS

27

75. AN EVENING O UT

Fiona followed her h u sband into the living room . " I t w a s a lovely evening," s h e said. " B ut you m u s t h ave spent about a hundred bucks . " " Not t h a t much, and an yway y o u d o n ' t h ave a birthday so ofte n . " Stan was checking h i s cash on the table . " You know, I 've still got a third as much as I h a d wh e n we went out. " "Then you c a n afford this," Fiona told h i m , picki ng u p two qu arters . " T h at p a y s b a c k w h a t I gave the h at check girl . " S t a n s m i l e d . " O kay. Now I 've got as many c e n t s in coins as I had dollars i n bills, and h alf as m any dollars i n bills as I had cents i n coi n s . " H o w m u c h did he h ave left? 76. A M ATT E R OF AGE S

"That's good of you and Pat, " said Bob, picking up a little photo on h i s colleagu e ' s de sk. " B ut who's the other m an?" Ted smiled. "That's my son , " h e replie d . " Robin. Maybe you'll meet him now he's b ack home for a wh i l e . " Bob examined the photo m ore closely. " I 'll t ake your word for it, but you must h ave started young ! " " Come on . " Ted chuckle d . "The cube o f Pat's age i s the differe nce betwe e n the square s of Rob's age and mine, so you see he's not so old . " " N o logic i n that." Bob shook h i s h e a d . "Just word s . " T e d sighed. " O kay. My age is exactly i n t h e s a m e ratio to Rob's as his is to Pat's, and I 've taken all three in complete ye ars . " Wh at were t h e three age s? 77. DROPPING IN ON LEN

Peter poi nted to the map. "We 're there, exactly midway between Ablin and Brent as the crow fli e s , and the same distance from Crowe , " he sai d . "All dead straight ro ads too. " " Sure . " J o e nodde d . " A n d L e n ' s place is j u st a third o f t h e w a y along t h e straight highway from Ablin to Crowe .

28

CHALLENGING MATHEMATICAL TEASERS

Of course we'll go through Abli n . We're only twenty-three m i l e s from him th at way." "That's right," Peter told him. "And e ach leg will be an exact n u m ber of mile s . " I n d e e d i t ' s an exact n u mber of m i l e s from Brent t o Crowe . H o w m a n y , then? 78. CARDS ON THE TAB LE

" I thought you were playing solitaire , " said Susan, look­ ing up from her book. "Why all the cussi ng?" "The table ' s too small." Len laugh e d . "Or m aybe the cards are too big, even though they are the regular two and one-qu arter by three and a half inches." " So the furniture ' s wrong ! " H i s wife smiled. "What's the game?" "Just an i d e a . I want to lay out a whole deck of cards, fi fty-two of the m , edge to e dge without overlapping or proj ecting beyond the table , " Len re pli e d . " B ut I can't fit t h e m all i n , and I ' m left with a few clubs and the same nu mber of he arts. Unfortunately it's the only rectangular table we h av e . " " T h at ' s right," S u s a n agre e d . " O n e i n c h longer t h a n i t i s wide . " " I know. " L e n was j otting down some figure s . " I 've still got eleven and one-qu arter inches of waste space, ar­ ranging the card s the best way I can . " Wh at were the d i m e n sions of the t abletop? 79. A B US RIDE

There were fewer than sixty of u s passengers in the bus when it left S a m ' s term i n u s . At B e nton , the first stop, a third of t h e passengers got off, and five got o n . At the next stop, again a third got off but only two got o n . At the third stop h alf got off and four came aboard . And at the fourth stop h alf got off but only one boarded the bus. Of course, nobody re-entered the b u s after alighting. Dorli ng was the fifth stop, the end of the ru n . Handing S a m my $ 2 . 5 0 fare as I stepped down , I re m arked on his rate s . " Only 50 cents per st age , and h alf price for kids," I said. " You'll go broke ! "

TE ASERS

29

The old chap gri n n e d . " It ' s enough for me. This trip the total take was exactly as m a n y dollars as the total n u m ­ b e r of passe ngers, and j u s t one-fi fth of them were k i d s a t t h e low rate . " How many adult passe ngers h a d there been? 80. YOU'RE SURE I F IN DOUBT Ted sipped his dri n k appreciati vely, glancing arou nd the cozy living roo m . "All the s a m e , as I recall it last t i m e I s a w y o u four y e a r s ago . How o l d a r e y o u r t w o kids n ow?" "Thre e , not two . Pam's had another m e anwhile . " Ch ar­ lie smiled. " I f you multiply their three age s you get ninety-six, but they add up to the n u m b e r of this h o u s e . " " You're sm art , " Ted t o l d hi m . " B ut that still d o e s n 't tell me for sure . " H i s friend chuckle d . " Sure it doe s . J u s t think." What do you m ake those age s? 8 1 . CRE E PY C RA WLIES " Slugs?" Tom chuckh�d . "I got rid of t h e m i n m y b ack­ yard, and none of those che mical killers ." "We 're infested with them," h i s friend told h i m . " W h at did you do?" " Simple and cheap," Tom repli e d . "I m ade a deal with our kids and the kids next door. E ach would get a d i m e for every s l u g he collecte d , but e ach w o u l d pay m e three cents e ach for all the slugs collected by the other kids in the gang. " Andy smiled. " Quite a sch e m e , but it doesn't sou nd so very cheap for you . " " I t was," T o m declare d . " I n fact i t cost m e an ave rage of exactly ninety-five cents for e ach of the gang. " How many slugs were collected? 82. FAM ILY N U M B E RS Doug was busy at the table . " S a y , Dad. I s Aunt E l s i e ' s phone n u m b e r 9638?" h e aske d . " A n d U n c l e Fre d ' s n u m­ ber 2 5 9 1 ?" "That's right, and don't forge t ours i s 8739 if you 're listing family n u m bers," replied Ste ve. " I s th at it?"

30

CHALLENGING MATHEMATICAL TEASERS

" No , but something I j u st figured out about those three n u m bers," the boy repli e d . " E ach of t h e m gives exactly the same re mainder when you divide by one special num­ ber I di scovere d . " W h a t w a s th at special n u m ber? 83. TOO MANY GIRLS

"What's new with the checke rs tourn ament at school?" Clem asked. " You did tell me you'd lost one game against some girl . " "All fi n i s h e d , D a d , " repli ed Jack. " E veryone h ad to play everyone else once . Mike and I were the only gu ys, and the girls all did better than either of us. Betwe e n us he and I scored only nine points i n all . " " So y o u played left-h ande d . " H i s father c h u ckled. " How were points counted?" " One for a win, h alf a point for a draw," the boy told h i m . " I t was fu n n y the girl s all got exactly the same n u m ber of poi nts . " H o w m a n y girls were in t h e tourn a m e nt? 84. WHAT? NO Z O B B LIES?

Mike missed the gurgling gluck-glucks and gru mphs of the zobbli e s when he went into h i s little pet shop. "What h appe n e d to those Ve n u s toad s?" he asked h i s wife . " Don't say you sold t h e m al L " " Sure d i d . E x actly a hu ndred of t h e m for exactly twe nty bucks," Susan re plied . "I knew the kids would go for them at those price s . Ninety-seven cents for five , sixty-seven cents for thre e , or si ngly a t a qu arte r eac h . " A m a z i n g ! O n l y a few ye ars ago man had n o t e v e n re ached t h e Moon ! How m a n y uf t h o s e creatures h ad S u s a n sold singly? 85. THE CENSUS

They've cou nted the c ats in Llanfair, Which n u m be r a third of a s q u are . If a q u arter were slai n , J u st a c u b e would re m ain . H ow m a n y , at least, m u st be there?

TEASERS

31

86. A WHIZ K I D " N oth ing t o it , e h ? Well, I 'd say ex actly a seventh o f the questions are really tough , so we 'll m ake a deal." Ch arlie chuckle d . " I ' ll give you 7¢ for e very one you get right. But you'll pay m e a p e n n y for the first mi stake , 2¢ for the second , 3¢ the third , and so o n . Any que stion you don't answer counts as a mistake . " John grin n e d . " Fi n e , D ad , " he said, re aching for the quiz sheet. " I t 'll cost you ! " I n fact, t h e boy answered more than three-qu arters of the questions correctly. but h e m ade only $ 1 .68 on the deal. How many questions were there? 8 7 . TIC KETS IN THE SWE E P "A kid fro m the school w a s h e re selling t ickets for your draw," said Susan . "I took thre e . " "That's fin e , M o m, " J ack told h e r . " B u t J i l l and I e ach got a ticket alre ady. There are a thousand in all, so let's see wh at your n u mbers are . " Susan h anded h i m h e r tickets fro m t h e dresser. " You see they're consecutive nu mbers. Let Jill keep the m , they'll b e safer with her." Jack ignore d th at re m ark ! " I t's fu n n y about t h e s e , " h e declare d . " T h e squ are o f your lowest n u m b e r a d d e d t o t h e squ are o f m in e i s the squ are of your high e s t , and that's also the squ are of your middle number added to the squ are of Jil l ' s . " Wh at were the five n u m bers? 88. PROGRE SS Ben put down h is gl ass. "I see you 're going to h ave a h igh-rise on TulIa Trail," h e said . " Will it bother you?" " We 're okay. I t ' s well up the road on the other s i d e , " J o e re pli e d . " T h e y 're the odd nu mbers . " " L ucky for y o u , " B e n commente d . " Wh at n u mbers are in volved?" Joe smiled. " I t's fu n n y about that. T h e y bought a block of adj oin i n g houses ru n n ing up to No. 43, their highest number. And I noticed that the sum of the n u m bers i n that block i s ex actly e q u al to the s u m of all the other

32

CHALLENGING MATHEMATICAL TEASERS

n u mbers that side below and above the ones they bough t . " What n u m bers were bought? 89. THREE TI M E S

S t a n walke d o v e r to where h i s son was writing a t the table . " B u sy?" h e aske d . " Not re ally, D a d , " Bill repli e d . " I j u st found something special about this n u m ber. If you move its fi nal pair of digits and put th e m i n front of the others you m ultiply the n u mber by three . " "That's o d d . " S t a n looked down a t t h e boy's working. " Will there be oth e r nu mbers like th at?" B ill nodded. " Sure , but you said 'odd,' and this is the s m allest odd n u m b e r that works t h at way for three times." W h a t was Bill's n u m ber? 90. A TOUCHING TALE

Tom stopped at the gate. "That's my little field," he said. " E xactly squ are , all fences i n good shape . " " Se e m s j u st wh at I want, b u t I 'll check th at fe nce," Andy repli e d . " W h at was the are a?" " You like teasers." Tom chuckle d . "We 're forty-eight fe et fro m that fe n c e , and the other gate in the side op­ posite us is one h u ndred and sixty-eight feet from it. If you want to touch that fe nce on your way to the other gat e , going righ t u p to it, you'll have to go at least forty­ eight fe et fu rt h e r than the direct route fro m where we are . " What w a s t h e area? 9 1 . F UN FOR SOME

"One of your proble m s?" M arth a asked, fi nding her hus­ band busy at t h e desk. "I don't know what you see i n the m . " Jack smil ed. " M aybe you will when you reach my age , " h e repli e d . " Now I 've fi gured o u t something on o u r ages. I f you write m y age after yours, the 4-figure nu mber will be twice the squ are of your age su btracted from three t i m e s the squ are of m i n e . " Just for fu n ! B u t w h at were their age s?

TEASERS

33

92. HE BROKE THE BAN K

M ary smiled, seeing the sorry m e s s on the rug. " S o that's t h e end o f your ugly o l d piggy bank," she s a i d . "Was there much i n it?" "Quite a bit, but it's fu nny the way it was," her h u s b an d replied. "The d i m e s , nickels and q u arters w e r e all pri m e numbers. And there were as many p e n n i e s as the n u m ber of d i m e s multiplied by the total nickels and d i m e s . " "What about qu arters?" M ary aske d . Stan grinned. " A good q u e stion , b u t figure it out your­ self. There we re 288 more pennies than qu arters, and no other money of course . " H o w m u c h was there i n all? 93. DO IT YOURSE L F

" H oly mackere l ! " exclai med S t e v e , viewing the litter on the floor as h e stood i n the doorway. "What's broke n?" " Getting on for fi fteen hu ndred little tile s , " Joe told him. " You re m e m ber m y two blackwood Chinese tables with little squ are tiles covering the tops? They h a d dry rot, so I decided to use all those ide ntical tiles for two new tables." " Sure , they were exactly the s a m e with squ are tops . " His friend nodde d . " Will the new t a b l e s be identical too?" Joe shook h i s head. " Squ are tops, but they were goi n g to be differe nt siz e s . One would h ave been twenty-three tile s , and I would h ave had e xactly the right total to cover both tops complete ly. But one d a m n e d table collapsed j u st as I fi nished stripping it, and all its tiles were sh at­ tere d . " Steve picked up s o m e fragments. " Fine ceramic, too bad," he com mented. " Wh at'll you do?" " Use what I h ave left," Joe repli e d . " Oddly enough they'll provide squ are tops for two s m aller tables without wasting any, though one will be rather s m al l . " How many tiles d i d he h ave left? 94. A B UG FOR THE B I RDS

The three sparrows perched expectantly at their re­ spective corners of the squ are bird fe eder. " D 'you see the big fat bug on th at flower, level with this tray?" asked Pi ppy. " It's still sixty-five inches from m e , so I 'll wait for it to come ne arer."

34

CHALLENGING M ATHEMATICAL TEASERS

Peppy, di agon ally opposite from her, re plied i n stantly. " Sure . It's only thirty-seven i nches fro m me, but it m ay come right on to this feeder tray if we're patient." The third bird was h u ngry . "Why wait?" he asked. " It's fi fty-one inches fro m m e , and I gue s s it's asleep . " T h e tray w a s fi at. What w a s i t s size? 95. A LUCKY NUMBER

" So you fell for the lottery too , " said R o n , glancing at the ticket. " B u t I don't see anything special about your n u mber. What m a k e s 1 09989 so luc ky?" J i m smil ed. " L ook agai n , " he repli e d . " I f you mu ltiply it by nine you get exactly the same digits in reversed order." Ron checke d . " T h at i s fu nny, but you can probably fi n d another 6-figure n u m ber t h a t works the same way with a different I-digit n u mber." Can you? 96. A MATTE R OF TI ME

Andy looked u p when h i s partner came i n . "I tried to get you yesterday afternoon , " he sai d . "They told me you'd gone out e arly, but you were n 't back when I left." "That's right. I did drop in on m y way home but you'd j u st gon e . " Greg replie d . " So here ' s a little teaser for you. When I c a m e back the m i nute h and was right on a m i n ute m ark exactly where the hour h and had been when I went out, and the hour hand was j u st two minutes ahead of where the minute h and had b e e n . " Wh at time had h e come b a c k t h at afternoon? 97. THE POSTER

" You know t h at old Swis s railway poster you gave m e , D ad?" said Peter. "Just n o w I folded it once , putting o n e corner on t o p of the opposite corner. T h e fold w a s exactly 136 centimeters long." Ch arlie smiled. " So wh at? I t was a big poster," he re­ pli e d . "I do re call it was 120 centi meters wide to fit a pro m otion display I was arrangi ng, but I d o n 't re me mber how long it w a s . " Wh at was its length?

TEASERS

35

98. AT THE CASINO

T h e y h ad been standing at the crowded roulette table , watching as the little white ball clicked its e rratic way to a final stop in e ach spin of the big black wh e e l . " O n e number has c o m e uJ? twice in the short t i m e w e ' v e b e e n here ," s a i d Alan . " DId y o u n otice?" Jim smiled. " Sure , but it could h appen quite ofte n , " h e replied. "With the regular 36 n u m be rs plus a si ngle zero as they h ave here , the odds are almost exactly 3 to 1 against getting the same number at least twice in the n u m be r of spins we've watch e d . " How m a n y spi n s was th at? 99. NO DIRE C T ROAD

" Ye s , I do my shopping at Poole , even though it i s eight miles from me as the crow fli e s , " said B eryl. "Just a m at­ ter of acce s s . " Sam n o d d e d . " T h a t ' s w h y we see y o u so often. But are n't Alton and Bray both ne arer to you?" he aske d . " I know the three vill age s are e q u al distances from e ach other, m aking a triangle . " " Sure , b u t that's what I m e a n , " Beryl replied. " I ' m only three miles from B ray and five miles from Alto n , but there ' s no direct road to either." How far apart were the three village s? 1 00. A NEST EGG

"That's right, a sort of trust fu n d . I started it on Mary's fi rst birthday with a deposit that was the product of her age and her mother's in dollars ," said Geoffrey. " Got the idea thinking about the coincidence that they h ave the same birthday. And every birthday since then I 've done the same, the new product of their age s . " Dick smiled. "A gre at idea, and it could be a very handy nest egg for the future. How much i s the fu n d now?" " I ' m not sure exactly, what with interest," Geoffre y replied. " B ut so far I 've deposited j u st $ 8 8 8 in a l l . " How o l d was her m o t h e r when M ary was born?

SO L U T I O NS TO TEASERS 1 . PE C K ING ORDER

Say there were x seeds. 8x - 76 4x - 20 2x - 4 Pip left Pe p left 9 ' Pop left 3 ' 27 2x + 8 Then, Pep took 9 . Pap took 2 +

8x - 130 . 8x + 32 I.e., ' 81 81 x + 4 9 - 2, whence x = 1 5 8.

8x + 3 2 = 81 They started with 1 5 8 seeds in the dish. So

--

2. FOR THE ONE THAT GOT AWAY

Say Clem bought y lures at x cents e ach. Then he sold them at x (x + 1 00)/ 1 0 0 = y 2 cents each. Hence, x 2 + 1 0 0x = 1 0Oy 2, so say x = 1 0z . The nce , (z + 5)2 - Y 2 = 2 5 . Tabulating for factors :

(z + 5) + ( z + 5) (z + 5)

y =

25 1

=

13 12

y =

y. =

Hence, z = 8, making x = 80, with y = 1 2 . S o C l e m bought 1 2 lures a t 80 cents apiec e , and s o l d t h e m a t $ 1 .44 e ach.

38

CHALLENGING MATHEMATICAL TEASERS

3. LONG O D D S

Say there we re ( x + 4) cards in all, including the four ace s . T h e n t h e c h a nce of drawing t h e four ace s was : 4 3 2 1 1 = . . . 1 00 1 · x + 4 x + 3 x + 2 x + 1 4 . 3 . 2 . 1 = 24. H e nce, as a very rough approxim ation, x 4 = 24000 (not more ). So, as a very rough approx i m ation, x 2 = 1 5 4 (not more). The nce , again as a very rough approxi m atio n , x = 12 (not more ). Now, 1001 = 7 . 1 1 . 1 3 , and all of (x + 4), (x + 3), (x + 2), and (x + 1 ) must be factors of 1 0 0 1 . The n, regarding that re quire ment concerning factors, none of (x + 4), (x + 3), (x + 2), or (x + 1 ) can be a multiple of 5. He nce we must have x = 1 0 . There were 1 4 cards on the table, including the four ace s.

4. LIKE FATHER LIKE SON

Age s : Sam ( l Ox + y), Mike ( l Oa + b). Then, ( x + y + a + b )2 = 2( 1 0x + y + l Oa + b ) - 1 and (x + y )2 - ( l Ox + y ) = (a + b )2 - ( l Oa + b). S a y : x + y = m , m aki ng l Ox + y = 9 x + m and a + b = n , m aking l Oa + b = 9a + n. Then, n 2 - (9a + n ) = m 2 - (9x + m ), whence (m - n)(m + n - 1) 9(x - a). Also, (m + n ) 2 = 2(m + n + 9x + 9a ) - 1 , s o (m + n )(m + n - 2) = 1 8 (x + a ) - 1 , which m akes (m + n ) odd, and therefore (m - n) odd . Also, since (m + n - 1 ) will be e ve n , (x - a ) is eve n . We had (m - n )(m + n - 1 ) = 9(x - a ) . Tabulate for possible e v e n ( x - a ) val u e s , noting t h at m < 19. =

x - a 9(x - a ) + n - 1 m - n m + n

m

n

m + n - 2 + n)( m + n - 2) m aking 1 8(x + a)

x + a x - a

w hence a nd s0 w ith a nd w ith S am's age M ike's age

18 3 19

24 3 25

6 3 7

18 1 19

12 3 13

5 2

10 9

8 5

5 35 36

17 323 324

11 143 1 44

2 2

18 2

8 4

= =

=

=

=

=

=

=

=

b

8 72

=

=

x y a

6 54

2 18

=

m

39

=

=

=

m

SOLUTIONS TO TE ASERS

=

=

=

2 3 0 2 23 2

4 36

-

-

-

-

-

-

6 2 2 3 62 23

36 1 37 -

-

-

-

-

-

-

-

11 8

14 11

17 323 324

23 575 576

18 6 -

-

-

-

Obviously the age s 23 and 2 can not be acce ptable , hence Sam was 62 years old, and M i ke 2 3 . Ann was 2 years old.

5. WHO S PI L L E D THE C O F F E E ? S a y x k n i v e s a t $2.96, y as the fi rst digit of the dollars and z as the l ast digit of the cents i n the a m o u nt . T h e n , 296x = 100000y + z + 70490, so B ut z < 1 0 , he nce z = 2 or 6 .

z =

4 k + 2 , say.

W i t h z = 2 , we h ave 2(37x - 1 2 500y) = 1 76 2 3 , w h i c h is i m possible because 3 i s o d d . 6, l e a d i n g to 3 7x - 1 2 5 0 0y = 8 8 1 2 , w i t h ge neral So z solution : x = 1 2 5 00t + 576, and y = 3 7t + 1 . =

1. 576, y Hence x The amount was $ 1 704.96 for 576 knives. =

=

40

CHALLENGING MATHEMATICAL TEASERS

6. SO FAR APART

Say Son i a ' s 7-digit n u m ber was x , initial digit not zero, and T e d ' s 4-digit n u m ber y . Then, 7x 3 = 3y 7 = 7 m3 "p 2 1 , say. The nce , x 3 = 7 m - 1 3 "p 2 1 , whe nce m = 3a + 1 , n = 3b . Also, y 7 = 7 m3 n - 1p 2 1 , whe nce m = 7c , n = 7d + 1 . S o , 3 a + 1 = 7c , e ntailing c = 3 k + 1 , a = 7 k + 2, say, and, 7d + 1 = 3b , entailing b = 7t - 2 , d = 3t - 1, say. Thence, m = 2 1k + 7, n = 2 1t - 6. So, x = 7 7k + 2 . 3 71 -2 . p 7 , Y = 73k + 1 . 331 - 1 . p 3. B ut, y < 1 0000, 331 - 1 � 9, 73k + 1 � 7, so p A l s o , 7 9 > 1 08, so we m u s t h a v e k = O. Then, since y < 10000, we must h ave t

=

1 or 2 .

=

1 or 2 .

T h e n c e i t i s trivial to show t h at we m u s t h ave : x = 2 7 . 35 72 = 1 524096, y = 23 . 32 . 7 = 0504. Soni a's n u m ber was 1 524096, Ted's was 0504. •

7. A L L TWOS

Say the individual amounts in one invoice were : x 2 dollars, x cents ; y2 dollars, y cents ; Z 2 dollars, z cents. And assume x > y > z. Then, 1 00x 2 + 1 0Oy 2 + 1 00z 2 + X + y + z = 22222, so x < 1 5 . Thenc e , (x + y + z ) < 4 0 . B ut w e must h ave ( x + y + z - 2 2 ) divisible b y 100, so (x + y + z ) = 22, whence z < 7. Also, (x 2 + y 2 + Z 2) = 222. Now try the possible values for z : z = 1 2 3 4 5 6 Z2 = 1 4 9 25 16 36 m aking x 2 + y2 = 22 1 2 1 8 2 1 3 206 1 9 7 1 8 6 . B ut 2 1 3 = 3 . 7 1 , 206 = 2 . 1 03 , 1 8 6 = 2 . 3 . 3 1 . So n o n e o f 2 1 3 , 206 a n d 186 c an be the sum of t w o square s , because e ach has a pri m e factor of form (4r - 1). Th e n : 221 = 1 02 + IF = 52 + 142, 2 1 8 = 72 + 1 32, 1 97 = 12 + 142. B ut we requ ire x + y + z = 22. 1 + 1 0 + 1 1 = 22, 2 + 7 + 1 3 = 22, 5 + 1 + 14 = 20.

SOLUTIONS TO TEASERS

41

There were the two invoice s , s o t h e respective amounts were : $ 1 2 1 . 1 1 , $ 1 00 . 1 0 , $ 1 . 0 1 for total $222.22 and $ 1 6 9 . 1 3 , $ 49.07, $4.02 for total $222 . 2 2 .

8 . THOSE SHARE S Say the apportionment was : x ,

VXy,

y ; with x

;?:

y.

Then say x = a 2c , y = b 2c , x y = abc . Thence , c (a 2 + ab + b 2) = 409. But 409 is pri m e , so c = 1 . Then (a 2 + ab + b 2) = 409 whence (2a + b )2 + 3b2 = 4 . 409. We m ay obviously ignore the fact th at 12 + 3 . 1 2 = 4, since (2a + b ) can not equal 1 here. So, say (2a + b ) = 2A, and b = 2B , whence A 2 + 3B 2 = 409. "At sight," 192 + 3 . 42 = 409, t h at being the unique representation since 409 is pri m e . Hence we must h ave A = 19, B = 4, whence a = 1 5 , b = 8. So x = 225, y = 64. The apportionment was 225, 120, and 64 s h ares. 9. YOU HAVE TO KNOW HOW

Say Tom's age was ( l Ox + y) years , Len's special digit z . Then, 70x + 7y = 1 00y + 1 0z + x , s o 69x - 93y = 1 0z . Say z = 3w , whence 23x - 3 1 y = l Ow . Since x , y , and w are integers, th at equation h as the general soluti o n : w = 31k - 7x , with y = 3 x - 1 0k . Obviously, since w e require w = 1 , 2 , o r 3 , with y < 10, we must h ave k < 3. With k = 2, we h ave w > 3. B ut , with k = 1, we h ave x 4, y = 2, w = 3 . Thence , z = 9, making T o m ' s age 42 years, L e n ' s digit 9 . =

1 0. SAVINGS

Say x dimes, y nickels, z q u arters , (z + 24) pennie s . Then, x (x + y ) = z + 24.

42

CHALLENGING M A THEMA TICAL TEASERS

If X = 2, the only even pri me n u m ber, then z would be even, which is i m possible since z � x . If z = 2, w e would have x (x + y ) = 2 6 = 1 · 26 o r 2 · 1 3 , but 1 is not a prime n u m ber, so that is un acceptable . So x and z are both odd pri m e s . He nce ( z + 2 4 ) i s odd, whe nce ( x + y ) is odd. B ut x � 2 , so y is even, he nce y = 2. T h e n z + 24 = x 2 + 2x , whe nce z = (x + 6)(x - 4). But z is pri m e , so we must h ave (x - 4) = 1 , whence x = 5. The nce z = 1 1 . S o the money box contained 1 1 q u arters, 5 dimes, 2 nickels and 35 pennies. 1 1 . A RANDOM MAILING

Say he picke d n card s , there being 20n cards i n the index. The n, successively he picke d nos. 1, 3, 6 , 1 0 , 15, etc., in the card index. n(n + 1 ) . So the fi nal card t h at he picked was no. 2 B u t that was in fact the last card in the index. So, since there were 20n cards i n all, n (n + 1) 4On , whence n = 3 9 . George picke d 39 cards. 12. NO COM PUTE R FOR TH IS

Say N = 1 0000x + y. The n, x 2 + y 2 = 30000x + 3y . Hence, (2x - 30000)2 + (2y - 3)2 = 900000009 = 9 . 1 7 . 5 882353 = 9 . 5882353 . (42 + 1 2) . W e require a repre sentation of 5 8 8 2 3 5 3 as t h e sum of two square s , say (a2 + b2). The n, ( 1 0 4 ) 2 + t 2 = (a 2 + b 2 )(4 2 + t 2 ) = (4a + b ) 2 + (a - 4b ) 2 . So we have : (4a + b ) = 10000, with (a - 4b ) = ± 1 , which has the solution : a = 2353, b = 588. Then 1 000000 0 1 = (23532 + 5882)(42 + 1 2) = (94 1 2 ± 5 88)2 + (2353 =+ 2352)2 = 88242 + 47052 or 100002 + 1 2.

SOLUTIONS TO TEASERS

43

There can be no represe ntation of 3 as the sum of two squ are s, so we h ave 900000009 = 264722 + 1 4 1 1 52 or 300002 + 32 • But x must have 4 digits, he nce (2x - 30000) *- ± 3000. So, 2x - 30000 = - 26472, with 2y - 3 = 1 4 1 1 5 , whe nce x = 1 764, y = 7059, making N = 1 7647059. NOTE : It i s not relevant i n this particular probl e m , but in fact 5882353 is a pri me n u mber, hence there was no alternative value for N. 1 3 . AT THE DINER

Until ide ntified by name a girl will be s hown as G, a m an as M. The seating arrange ment includes two seque nce s : Ron, G , M , Joan ; and Ann, M , G , M , G , Pam's husband. But the girl to left of Pam's husband was not Pam, so must h ave been Joan. Hence, sitti ng around the table, they were : Ann, Ron, Pam, M, Joan, Pam's husband. But Steve sat on the right of the girl who sat on Harry's right. He nce the arrangement must h ave bee n : Ann, Ron , Pam, H arry, Joan, Ste ve. Steve was Pam's husband. 14. THEY DON'T COME SINGLY

Check for x dollars and y cents, i . e . , ( 1 00x + y ) ¢. E ach incorrect e ntry was for ( 1 0 Oy + x ) ¢ , so ( 1 00x + y ) - (20Oy + 2x ) = ± 1 333, whence 98x - 199y = ± 1 333, with ge neral integral solution : x = 1 99k - 4 0 x = 1 99k + 40 or y = 98k - 13 y = 9 8k + 13 But we require x < 1 0 0 , he nce x 40, y = 1 3 . Check was for $40. 1 3 . 1 5 . WHO GLUBS GL YGONS?

Say they gathered x glygons, x being " n e arly 1 0 0 0 . " F l a b left (3x - 12)/4.

44

CHALLENGING MATHEMATICAL TEASERS

Finally, Flub left (8 1x - 804)/256. So, (81x - 804)/2 56 = 4y , say. Then 8 1 x - 1 0 24y = 804, with integral solution: x = 1 024k - 28, y = 8 1k - 3. B u t we re q uire x "ne arly 1 000," so x = 996, with y They gathered 996 glygons.

=

78.

1 6. THREE IN A ROW

Say the age s were x, y and z ye ars ; all less than 10 years. Then, 1 00x + l Oy + z = 32(x + y + z), whe nce 6 8x - 22y = 3 1 z , so z is even. Say z = 2w , making 34x - 1 1y 3 1w , with w 1 , 2, 3, or 4. =

=

Dividing by 1 1 , we h a v e x = 1 1k - 2w , y 34k N o w 2 w < 1 0 , and x < 1 0 , he nce 1 1k < 2 0 , so we h ave k = 1 . The nce , x = 1 1 - 2w , y = 34 - 9w . B ut y < 1 0 , s o w > 2. Also we must h ave 9w < 34, so w < 4. He nce w = 3, with x = 5 , y = 7, z = 6. =

-

9w .

The age s , i n the order in which Ray wrote them down , were 5 , 7, and 6 years. 1 7. A SWITCH

Say the n u m ber was ( 1 0000y + x), with x < 1 0000, y < 1000. W i t h the t w o parts " switche d " it becom e s ( 1 000x + y), so 1 000x + y = 2( 1 0 00 Oy + x ) + 1 , whe nce 998x - 1 9 999y = 1 . Dividing through b y 998, (39y + 1 )/998 must b e a n integer. I n order to obtain a re sult such that (y + t )/998, say, will be an integer, we c arry out the elementary arithmetic routine for fi nding the H . C . F . (i.e . , Highest Common Factor) of two n u m be rs : 1 3 9 998 2 5 2 3 975 2 1 6 23 1 16 14 "2 -7 3 6 -1

SOLUTIONS TO TEASERS

45

1 1 1 1 1 17 + + + = + 1 1 25 "2 3 435 So we multiply (39y + 1 )/998 by 435, making 1 6965y + 435 . . , whIch m ust be an mteger. Thence, 998 -y + 435 . ' x = 8717. IS an I. nteger, so y = 435 , rn a k mg 998 The phone number was 435- 8 7 1 7 . Then the conti nued fraction

----

1 8 . J U S T TRIANGLE S Say sides of one tri angle were (x + y), (x + z), (y + z ) inches. Then are a was vxyz(x + y + z), perimeter 2(x + y + z ) , whe nce 9xyz = 16(x + y + z). 3 . 3 . 3 27 xyz 16 Now = - . But, = x + y + z 9 3 + 3 + 3 9' so taking x ;?: y ;?: z , no solution i s possible with z > 2 . S a y z = 1 . Then 9xy - 1 6x - 1 6y = 1 6 , s o (9x - 1 6)(9y - 1 6 ) = 400. Taking factors for integral x and y : 9x - 1 6 = 200 20 9y - 1 6 = 2 20 x = 24 4 y = 2 4 So, sides : 26, 25, 3 and 8, 5 , 5 . Say z = 2. Simil arly w e h a v e (9x - 8)(9y - 8) = 208, giving x = 24, y = 1 , with sides 26, 25, 3. Hence there were only the two different tri angles, with sides 26, 25, 3 inches and 8, 5 , 5 inches. 1 9 . AN ANCIENT PROB LE M We h ave i ntege rs A , B , C such that (A + B + C ) i s a square , and each of (A + B ), (B + C), and (A + C ) is also a squ are . The gre at mathe matician Diophantos devel­ oped the ge neral i ntegral solution for this proble m about 1700 ye ars ago. Say A + B + C = (x + y )2, (A + B ) = x 2 with C = (2xy + y 2) and (B + C ) = (x - y )2. Then, B = (x 2 - 4xy ) , with A = 4xy , C = (2xy + y 2).

CHALLENGING MATHEMATICAL TEASERS

46

We re q uire (A + C) to be squ are , so (y 2 + 6xy ) Z2, say, m aki ng (3X )2 + Z 2 = (3x + y )2, which has the general integral solution : x (m2 - n2)k , y 6n2k, Z 6mnk, m and n being any i ntegers, k a common factor. =

=

=

The nce : A

=

C

=

=

24n2(m2 - n2)k 2, B (m2 - n2)(m2 - 25n2)k2, 1 2n2(m2 + 2n2)k 2, A + B + C = (m2 + 5n2)2k 2. =

But for our particular problem we re quire values within the l i m its of human age s . Hence, re the expre ssion for B , we must have m > 5n . The n , observing that (A + B + C ) < 13, quick trial of a very few (m, n) values gives the only acceptable solu­ tion with m 7, n 1, k 6. Thence A = 32, B 32, C 17. Age s : B e s sie 32 years, J o e 32, Peter 1 7 , Sally 9. =

=

=

=

=

20. S I M I LAR B UT D I F F E RENT

Say the tri angles had sides ax, ay, az and bx, by, bz. And say ax bz, az by. Then, z ax/b by/a, so x/y b 2/a 2. H e nce, x b 2k, y a2k, z abk, say. The nce : ax a b 2k bx b3k a2bk ay a3k by a b 2k az = a 2bk bz So, (a 3 - b 3 )k 1 9 , which is pri m e , whence k 1 , m aking a 3, b 2. =

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

The sides were : 12, 18 and 27 inche s , 8 , 1 2 and 1 8 inches.

2 1 . J UST FOR K I D S

S t a n bought 1 2x a t 1 2x ¢ , 1 2y a t 1 2y ¢ , 1 2z a t 1 2z ¢ , with x > y > z say. Total cost 1 44(x 2 + y 2 + Z2)¢ for 12(x + y + z ) h ats at average price 50¢. H e nce, 1 44(x 2 + y 2 + Z 2) 600(x + y + z ), so ( 1 2x - 2 5 )2 + ( 1 2y - 2 5 )2 + ( 1 2z - 2 5 )2 =

=

1875.

SOLUTIONS TO TEASERS ( 1 2z - 25)2 < ( 1 2y - 25)2 < ( 1 2x - 25)2, so ( 1 2z - 25)2 < 6 2 5 , hence z < 5 . Also z

>

47

2.

4, ( 1 2x - 25)2 + ( 1 2y - 25)2 = 1346 With z 352 + I F, making x 5, y 3, which is not acceptable because we re quire y > z. The n, with z = 3, we have y 4, and x 5. =

=

=

=

=

=

Stan bought 6 0 a t 6 0 ¢ , 48 a t 4 8 ¢ , 36 at 3 6 ¢ , a total of 1 4 4 hats.

22 . THE PATIO

The ancient Heronian form ula for the are a of any tri angle is: A = ys (s - x )(s - y )(s - z), where the sides are x, y, z, and 2s x + y + z. =

Substituting for s , this m ake s : [2yz + (x 2 - y 2 - z2)][2yz - (x 2 - y 2 - Z2)] 1 6A2 (2YZ )2 - (x 2 - y2 - z2)2. =

=

Here we h ave x2 1 9 6 , y2 97, Z2 41, whence 1 6A 2 1 2 544, he nce A 28. The are a of the u n paved tri angle was 28 squ are feet. =

=

=

=

=

23. SQUARE S AND SQUARE S

Age s : Sam A , wife B , J ack y , house n u m ber x , all integral. Then x (y + 1 ) 2A 2, and y (x + 1 ) 2B 2, which entails 2a 2 c 2 - 1 , and y = 2b2 d2 - 1 , say, where a , b, x c , d are integers . 1 , d2 - 2b2 1. Then, c 2 - 2a 2 So t h e pairs (c , a ) a n d (d, b ) are particular pairs i n the i ntegral solution of p2 - 2Q 2 1. =

=

=

=

=

=

=

=

In that Pe ll equation all succe ssive P, Q values are : P 1 3 17 99 etc. Q 0 2 1 2 70 etc . , P a n d Q obeying t h e relation U n + 2 6 U n + 1 - U n. =

=

=

48

CHALLENGING MATHEMATICAL TEASERS

Tabulate the first few values with corre sponding x or y value s : p = c or d = 1 3 17 99 etc. Q = a or b = ° 2 12 70 etc. x or y = ° 8 288 9800 etc. B ut y is the son's age , so y = 8, with x = 288, m aking A = 36, B = 34. Sam's age was 36 ye ars , h i s wife ' s , 34 years.

24. WRONG B UT RIGHT

Say the three a m o u nts were x cents, y cents, z cents, with x � y � z . Then i n doll ars they were x/100, y/100, z/100. Then the sum was (x + y + z )! 1 0 0 , the product xyz/1 000000. B ut the total was $ 5 . 7 0 , so x + y + z = 570, xyz = 57 . 1 05. Say z = 1 0 0 , m aking xy = 57000, and x + y = 470, so x (470 - x) = 57000, whe nce x 2 - 470x + 57000 = 0, m a king x = 235 ± v55225 - 57000, which is i mpossible . He nce z > 1 0 0 , and obviously z < 1 90. Now, z m ust be a factor of 5700000 = 25 . 3 . 55 . 19, so z m ust be one of 1 1 4, 1 2 0 , 1 2 5 , 1 5 0 , 1 5 2 , 160. Solving the corre sponding qu adratic in x for e ach possi­ bility, i n the same way as for the case of z = 100, we d erive non-rational values for x in the cases of 1 1 4, 120, 1 5 0 , and 1 60 . B ut with z = 1 2 5 , we h ave x = 285, y = 1 6 0 ; with z = 1 6 0 , we h ave x = 285, y = 1 2 5 , which is i m possible si nce we re q uire y � z . So x = 2 8 5 , y = 1 6 0 , z = 1 2 5 . Price s : $ 2 . 8 5 , $ 1 .60, $ 1 . 2 5 .

2 5 . THE STA M P

(3 + 9 + 1 0 + 1 2 + 1 3 + 14) = 6 1 == l(mod 3) The 5 st amps used at the airport totale d a mu ltiple of 3 kuks. He nce the re m aining st a m p had to be of value == l(mod 3), so was for 1 0 kuks or 1 3 ku ks. Say 1 3 ku k s : (3 + 9 + 10 + 12 + 14) = 48 but no combination will give 1 6 . Say 1 0 kuks : ( 3 + 9 + 1 2 + 13 + 14) = 5 1 and 3 + 1 4 = 1 7 , 9 + 1 2 + 1 3 34. =

=

1 6 . 3,

=

1 7 . 3,

SOLUTIONS TO TEASERS

49

So the postc ards took stamps for 17 kuks and 34 kuks . Andy retained the 1 0-kuk stamp. 26. A TALE O F TWO GUYS

Say the le ngths of the wire s were x and (250 - x) fe et, with x > (250 - x ) , and the e n d of the shorter wire was y feet fro m the pole at the ground. Then, c 2 - 2a2 = 1, d2 - 2b2 = 1 . s o V784 + (y + 150)2 + V784 + y2 = 250

(A)

But [784 + (y + 150)2] - [784 + y2] = 300( y + 75), so V784 + (y + 1 50)2 - V784 + y2 = 6(y + 7 5)/5

(B)

Subtracting (B) from (A), V784 + y2 = (400 - 3y )/5, whence y = 45, m aking x = 197. So the lengt h s of the wire s were 1 9 7 and 53 fe et. 2 7 . J UST JUNK

Say the number was N, with digits 1x 1y 1z 1 i n that order. Then N == - l (mod 7) == - l (mod 1 1 ) == - l (mod 13), so N == - l (mod 1 0 0 1 ). N = 100000x + 1 000y + 1 0z + 1 0 1 0 1 0 1 , s o 100000x + 1 000y + 1 0z == - 1 0 1 0 1 02 == 908(mod 1 0 0 1 ), whence 1 0 ( 1 0 000x + 1 00y + z ) = 1 0 0 1k + 908, say. That requires k = l Ot + 2, say. Thence , 1 0000x + 1 0 0y + z = 1 0 0 l t + 2 9 1 . T h e n dividing through by 1 0 0 , a n d since t i s obviously less than 100, we must h ave t = z + 9 . Hence, 1 0000x + 1 0 0y + z = 100 1z + 9300, so 1 00x + y = 1 0z + 93. B ut y < 10, so the 2nd digit of ( 1 00x + y ) must be zero. He nce z = 1 , making x 1 , y = 3. Then N = 1 1 1 3 1 1 1 . =

28. A S I M PLE ROUTINE Say the number was ( 1 000x + y), where y < 1 000, x < 1 0 0 . T h e n , y2 - x 1 000x + y , whe nce (2y - 1 )2 = 4004x + 1 . S o , (2y - 1 )2 == l ( m o d 7) == l ( m o d 1 1 ) == l(mod 13) and is odd. Then, (2y - 1 ) == ± l ( mod 14) == ± l(mod 13) == ± l ( mod 1 1 ). =

CHALLENGING MATHEMATICAL TEASERS

50

Say (2y - 1) = 1 4u + a = 1 3v + b, which e ntails u 1 3k - a + b, m aking (2y - 1 ) = 1 82k - 1 3a + 14b == ± 1 or ± 27(mod 1 82). The n say (2y - 1 ) = 1 82m + c = 1 1n + d , which entails m = 1 1t - 2c + 2d , where c = ± 1 or ± 2 7 and d = ± 1 . T h at m ake s (2y - 1 ) = 2002t - 363c + 364d , whe nce (2y - 1 ) == ± 1 , or ± 1 5 5 , o r ± 5 73, or ± 727 (mod 2002). B u t (2y - 1 )2 < 4004x + 2 , so (2y - 1 )2 < 396398, m aki ng (2y - 1 ) < 63 1 . =

S o there re main 3 possibilitie s : (2y - 1 ) 1 1 5 5 573 making y = 1 78 287 with x = 0 6 82 We may disregard the first pair of values, so le aving the two n u m bers 6078 and 82287. U ncle Fre d ' s n u m ber was 82287. =

29. C O M M E M ORATION STA M PS

x sets: 3 at 4¢, 2 at 5 ¢ , 5 at 1 0 ¢ . y s e t s : 7 a t 4¢, 3 at 5¢, 3 at 1 0 ¢ . z sets: 1 a t 4 ¢ , 5 a t 5 ¢ , 4 at 1 0 ¢ . Say t here were w stamps of e ach denomination. Then, 3x + 7y + z w , 2x + 3y + 5z = w , 5x + 3y + 4z = w , whe nce z 3x . So, 4y 1 2x - x 1 1x , he nce y = 1 1k , say, and x = 4k , m aking z 1 2k . Total sets re m aining would b e 27k ; so, since " about 50" sets re m ai n e d , k = 2. T h e n x = 8 , y = 22, z = 24, and w = 202. So 606 stamps re m ained. =

=

=

=

=

30. ALMOST NE IGHB ORS

Say Ke n ' s n u mber was x . S u m o f n u m bers 1 t o 7 3 7 4 . 73/2 = 270 1 . S u m o f n u m bers (x + 1 ) t o (x + 73) 73x + 270 1 . S u m o f n u mbers 1 t o (x + 73) = (x + 73 )(x + 74)/2. =

=

SOLUTIONS TO TEASERS

51

So, 73x + 5402 = (X 2 + 1 4 7x + 5402)/4, whe nce x 2 - 1 4 5x - 16206 = 0, so x = 2 1 9 . Ke n's n u m be r was 2 1 9 . 3 1 . RE AL E STATE

Are a of any tri angle is vs (s - a )(s - b )(s - c ) , where the sides are a, b, c , and 2s = a + b + c. Say sides (x - 7), x , (x + 7) yard s, making s = 3x/2, and are a = x v3(x2 - 196)/4. For i ntegral values we must h ave 3(x 2 - 196) a perfect square , so say x 2 - 3y 2 = 1 9 6 = 22 . 72, are a = 3xy/4. Neither A 2 - 3B 2 = - 7, nor A 2 - 3B 2 = 7, can h ave an integral solution, so say x = 1 4K, y = 1 4Y, m aking X2 - 3P = 1 . Tabulate for succe ssive X , Y values (both X and Y conform to the relation U n + 2 = 4U n + 1 - U n , i . e . , 26 = 4 · 7 - 2): X Y x y 3xy/4

= 1 = 0 = 14 = 0 = 0

2 1 28 14 294

7 4 98 56 4116

26 15 364 210 57330

etc. etc. etc. etc. etc.

But 1 2 acre s is 58080 squ are yards, so the are a must h ave been 5 7330 squ are yards, with sides 357, 364 and 3 7 1 yards. 32. GRE ETINGS

Age s : Uncle Tom X ; M ark and Judy, (Y + 1 ) and Y. The n, (Y + 1 )3 - ya = X2 + 1, so 3P + 3Y = X2. Say X = 3Z . Then, (2Y + 1 )2 - 3(2Z )2 = 1 , with successive solutions : 2Y + 1 = 1 7 97 etc. Z = 0 2 28 etc. NOTE : Both (2Y + 1 ) and Z obey the re lation U n + 2 = 1 4 U n + l - U n. Here we must obviously have Z = 28, m aking X = 84 with Y = 48. Uncle Tom was 84 ye ars old.

52

CHALLENGING MATHEMATICAL TEASERS 33. AWAY FROM IT ALL

x

x

2P + Q = 1 80°, so cosQ = - cos2P = 1 - 2cos2P. B u t we have cosP = (2x 2 - 900)/2x 2 = (x 2 - 450)/X 2 and c o s Q = (2x 2 - 196)/2x 2 = (x 2 - 98)/x2. Thence, X4 - 949x 2 + 202500 = 0, so x 2 = 324 or 625. We re quire positive values, so x = 1 8 or 2 5 . But, w i t h x = 1 8 we w o u l d h ave t h e angle P gre ater than 90°, m aki ng 2P > 1 80°, which is not acceptable since the roads to Carlow and Lawanda both led off on the same side of the B rill/Sefton road. He nce we m ust have x = 2 5 , m aking the four villages all 2 5 kilometers from Mike 's gate s . 34. DOUBLE THE O D D S

Say there were x gre e n m arbles , y re d . After 2 gree n s h a d b e e n drawn the ch ance of drawing a t hird gre e n was (x - 2)/(x + y - 2), with odds y to (x 2) against the e ve nt. The odds against drawing a red i n itially had be e n x to y agai nst. He nce y/(x - 2) = 2x/y , whence 2X 2 - 4x = y 2. Setting y = 2z , t h at beco mes (x - 1 )2 - 2Z2 = 1 . All integral solutions m ay b e (x - 1 ) = 1 Z = 0 with m aking x = 2 with y 0 =

tabulated a s : 3 17 99 etc. 2 12 70 etc. 4 1 8 100 etc. 4 24 140 etc.

There were " about forty," so Jim had 1 8 gree n and 24 red m arble s .

SOLUTIONS TO TEASERS

53

35. B O UNTY H UNTERS

Say x boys killed y e ach, (50 - x) girls killed z e ac h . Then 2 5 y - z(50 - x ) = 1 9 , and 25z - xy = 1 1 . Elimi nating z , 625y - (xy + 1 1 )(50 - x ) = 475, so yx 2 - (50y - l 1 )x + 625y - 1 02 5 = O . Thence, 2yx = (50y - 1 1 ) ± V(3000y + 1 2 1 ). Say 3000y + 1 2 1 = k2, which h a s the general integral solution k == ± 1 1 or ± 239 (mod 750). Hence, k = 11 with y = 0 m aking x = -

239 19 31

511 87 22

etc. etc. etc.

But, at most, xy i s " around 1000." Hence we must h ave x = 31, with y = 1 9 , z = 24. There were 1 9 girl s, who killed 24 e ach ; and 31 boys who killed 19 e ach . 36. THE M URAL

We h ave, say: xy = 2(w + z), and wz = 2(x + y ) where x , y , z , w are integers. And w e m a y assume x 2: y , with x 2: w, Y 2: w, Z 2: w (i.e . , none of x, y, z is less than w). Combining the e q u ations, (x - 2)(y - 2) + (z - 2)(w - 2) = 8 , so we must h ave w < 5. We can also combine the e q u ations to give xyw - 4x - 4y = 2W 2, whence (wx - 4)(wy - 4) = 2w 3 + 1 6 . S a y w = 1 . Then, ( x - 4)(y - 4) = 1 8 = 1 8 . 1 , 9 . 2 , or 6 · 3 ; whence x = 22 or 13 or 1 0 with y = 5 6 7 and z = 54 34. 38 Similarly, if w = 2, x = 10 or 6 4 with y = 3 10. and z = 13 If w = 3, x = 13 or 6 3 with y = 2 6. and z = 10 I f w = 4, x = 10 or 4 4 with y = 2 and z = 6 4.

54

CHALLENGING MATHEMATICAL TEASERS

Hence we h ave 7 different pairs : 54 by 1 , 22 by 5 ; 38 by 1 , 1 3 by 6 ; 34 by 1 , 1 0 by 7 ; 1 3 by 2, 1 0 by 3 ; 1 0 by 2, 6 by 4 ; 6 by 3, 6 by 3 ; 4 b y 4, 4 b y 4 . 3 7 . A WOMAN'S JOB

S ay B ill could do l/x of the j ob per minute, Ann 1/12y of the j ob. To do the lot alone Ann would t ake 1 2y m inutes. So Bill worke d alone for 5y minute s , le aving (x - 5y )/x for Ann to do. To do that, Ann took 1 2y (x - 5y )/x minute s . Together t h e y would h ave take n 1 2xyl(x + 1 2y ) minutes to do the lot, i n which time B ill alone would h ave done 1 2yl(x + 1 2y ) of the j ob . So, 1 2yl(x + 12y) = ( x - 5y )/2x , w h e n c e x = 20y . I n fact, they took 5y + 1 2y (x - 5y )/x m inute s , i . e . , 7x/ 1 0 minute s . Together t h e y w o u l d h ave take n 1 2xyl(x + 1 2y), i . e . , 3x/8 mi nute s . So, 7x/ 1 0 - 3x/8 = 5 2 , whe nce x = 160, with y = 8. Their total time was 1 hour and 52 minute s . 38. E A C H T O E A C H

Say the n u mbers of people involve d in t h e exchange of cards for the three succe ssive ye ars were z, y, x, where x > y > z. Corre sponding n u mbers of c ard s : x (x - 1 ) , y (y - 1 ) , z (z - 1 ) . So, (x 2 - x ) - (y2 - y) = 8 4 , i . e . , (2x - 1) 2 - (2y - 1 ) 2 = 4 . 8 4 = 4 . 22 . 3 . 7. Tablulate for factors : 84 42 28 2 1 14 1 2 [(2x - 1 ) + (2y - 1 )]/2 [(2x - 1 ) - (2y - 1 )]/2

1

2

3

4

6

7

85 44 31 25 20 1 9

(2x - 1 ) (2y - 1) x

83 40 2 5 1 7

8

5

43 - 16 13 - 1 0

42 - 13 9 y S i m i l arly we get "possibles" for y and z as : y = 43 16 13 1 0 z = 42 13 9 3

-

3

SOLUTIONS TO TEASERS

55

B ut t h e value o f y m ust b e t h e same in each case. So, x = 1 6 , y = 13, Z = 9. That l ast X m as 16(16 - 1 ), i.e., 240, cards were exchange d .

3 9 . THE J E WE L B OX

(x 2 + 39)/2x . y2 = 52 - (x - Z )2 = 82 - Z 2, so z Z2 = 82 - y 2 = 132 - (x - y ) 2, so Y = (x 2 - 1 0 5 )/2x . The n, y2 = (x 2 - 1 0 5 )2/4x 2. But, y 2 = 64 - Z 2, so y2 = 64 - (x 2 + 39)2/4x 2, he nce , (x 2 - 1 0 5 )2 256x 2 - (x 2 + 39)2, 4 1 or whe nce X4 - 194x 2 + 6273 0, with solution x 2 1 53. B ut, from the e x pre ssion for y , x 2 > 105. Hence we must h ave x 2 1 53. So the top of the box h a d are a 1 5 3 squ are c m . =

=

=

=

=

40. PETER'S PENNI E S

Say Peter h a d m2 pennies. Totals for hexagon formation are succe ssively 1 , 7, 1 9 , 37, etc. , from which we see that the total m2 = 3n2 - 3n + 1 , corresponding to n pennies per side in the hexagon , m a n d n being w h o l e n u m bers. Then, 6n 3 ± V3( 4m 2 - 1). For integral values 4m2 - 1 = 3k2, say. H e nce, (2m)2 - 3k 2 1. =

=

56

CHALLENGING MATHEMATICAL TEASERS

Succe ssive values in this e q u ation are : m = 1 13 1 8 1 etc . , k 1 1 5 2 0 9 etc., both m and k obeying the rule Ur = 1 4 U r- 1 - Ur- • 2 B u t 1 32 = 1 6 9 , and 1 8 1 2 = 3276 1 , so obviously Peter had 1 6 9 pe nnies o n the table. Then 3276 1 pennies would be re quire d for the next bigge r case, so that would entail 32592 m ore . =

4 1 . SO VERY SIMPLE

The number was x, with (n - 1 ) digits, special digit y. Then 29x = l Ox + ( 1 0 n + 1 )y , whence ( l o n - 1 )y = 1 9x , so l O n == - l (mod 1 9). All to (mod 19): 1 02 == 5 , 1 03 == - 7, 1 0 4 == 6 , l O s == 3, 1 0 6 == - 8 , 1 0 7 == - 4, 1 0 8 == - 2 , 1 0 9 == - 1 . So we have n = 9, he nce x h a s 8 digits. Then 1 9x = 1 00000000 1 y , x = 52631 579y . B u t ou r x h a s 8 digits, s o y = 1 , x = 52631 579. Mike ' s number was 5263 1 5 79, his " s pecial digit" was 1. 4 2 . POWE R PLAY

Say x 2 = 2y3 = 3z s = 22a . 32b , he nce x = 2 a . 3 b . T h e n, y3 = 2 ( 2a - I ) 32b , so 2a - 1 = 3u , say, whe nce a = 3m - 1, b = 3n , say, n making y3 = 2 (6m- 3 ) . 3 6 , hence y = 2( 2 m - I ) 32 n . n ) m The n x 2 = 2 (6 - 2 . 3 6 = 3z s , n so Z S = 2 (6m -2 ) . 3 (6 - I ) , hence m = 5k + 2, n = 5t + 1 . = 2 (6k + 2 ) . 3(61 + 1 ) , The n w e h ave z I wi th x = 2 0Sk + S) 30 S + 3 ) , + + 0 ) 1 0 and y = 2 0k 3 3 0 2 ) . For m i n i m al positive values k = t = 0, m aking x = 2 s ' 33 = 864, y = 23 . 32 = 72, z = 22 . 3 = 1 2 . •

•

•

•

The n e x t gre ate r v a l u e s e ntail k = 1 , t = 0, m aking x = 220 . 33, obviously unacceptable as a house nu mber. So the n u m bers were : B ruce 864, J ane 72, Tom 1 2 . 43. T H E COLLE CTOR

59x + 1 9 9y + 287z + 344( 1 0 0 - x - y - z) whence 285x + 1 4 5y + 5 7z = 24400, so 5 7(5x + z ) + 1 4 5y = 24400.

=

1 0000,

SOLUTIONS TO TEASERS

57

Then, since x , y, z are integers, (24400 - 145y) must b e divisible by 57, he nce y = 5 7 k + 1 3 w i t h k any integer. Now, y < 1 0000/ 1 9 9 , so y < 5 1 . Hence we must h ave k = 0, with y = 13. T h e n 5x + z 395, so x + z = 395 - 4x . But (x + z ) < 87, so 4x > 308, whe nce x > 77. Also, re ( 5x + z), x < 79. So x = 78, m a king z = 5 . He nce Walt bought: 78 a t $0.59, 1 3 at $ 1 .99, 5 a t $ 2 . 8 7 , 4 at $3.44. =

44. A MATTER O F SQUARE S

Say the number had digits x, y, z in that order. Then, x 2 + y 2 + Z2 = ( 1 00x + l Oy + z )/2, so z must be e ve n . Say z = 2w , m aki ng x 2 + y 2 + 4w2 = 5 0 x + 5 y + w , whe nce ( 5 0 - 2X)2 + (2y - 5)2 = 2 5 2 5 - 4w (4w - 1), with w < 5. S e t X = 50 - 2 x and Y = 2 y - 5 , where X < 5 0 , Y :5 1 3 . Now consider t h e possible values for w . Note t h at if N = kt 2, k being quadratfrei (i.e . , no squ are factor), then N cannot be the sum of two integral squares if k has any factor of the form (4n - 1 ) . = 0, X 2 + P = 2525 = 502 + 52 = 3 4 2 + 3 7 2 = 262 + 432, all being un acceptable. w = 1 , X2 + y2 = 2 5 1 3 = 3 5 9 ' 7 , impossible re 7. w = 2, X2 + P = 2469 = 823 ' 3 , impossible re 3. w = 3 , X2 + y2 = 2393 = 322 + 372, un acceptable . w = 4, X2 + P = 2 2 8 5 = 4 5 7 · 5 = ( 2 P + 42)(22 + P) 382 + 292 = 462 + 1 32. But Y :5 13, so we must h ave X = 46, Y = 1 3 .

w

=

The nce x

=

2, y

=

9, z

=

8, m aking the n u m ber 298.

45. THE ONLY TRE E

House n u m ber 2n , the even n u m bers going up to 2m . Then sum of even n u m bers above Ste v e ' s was : [m(m + 1 ) - n (n + 1 )], and sum of eve n n u m bers below his was : n (n - 1). So m(m + 1 ) - n(n + 1 ) = n(n - 1 )/2. (A) Whence (6n + 1)2 - 6( 2m + 1)2 = - 5

58

CHALLENGING MATHEMATICAL TEASERS

Now, 12 - 6 · 1 2 = - 5 , so all integral solutions of (A) are given by: 6n + 1 = ± (a ± 6b), 2m + 1 = ± (a ± b), the " i ntern al" + and - signs i n agre e m e nt, where a2 - 6b2 = 1. (B) The two s m allest i ntegral solutions of ( B ) are : (a, b ) = ( 1 , 0) and ( 5 , 2), so all further successive (a, b ) pairs are given by: a r+ = 1 0a r+ l - a n b r- = 1 0b r+ 1 - b r •

2

2

We now tabul ate for succe ssive (a, b ) pairs, observing that we re q uire n < 5 0 : a = 1 5 5 49 49 485 etc. 2 0 198 etc. b =O 2 2 20

2m + 1 = 1

6n + l = 1

7 3

n = 0

1

17 7

71 29

1 69 69

703 287

etc. etc.

28 1 1 7 etc. 34 143 etc. So t here were 34 e ve n n u mbers, Steve 's being 56.

m =0

1

46. IT MAKES SENSE

H e re we h ave calculations using a scale of notation that i s not our u s u al d e nary scal e . I n solving the proble m it is conve nient to use the d e n ary scale, however. Say the calculations were in scale-n notation, i . e . , to base n. T h e n w e have (n2 + 2n + 1 ) + (2n2 + n + 2 ) = 43 ·3, so n2 + n - 4 2 = 0 , whe nce n = 6 . He nce 1 1 1 m e ans ( 3 6 + 6 + 1), i . e . , 4 3 i n scale 1 0 , and 1 1 m e a n s (6 + 1 ) , i . e . , 7 in scale- l 0 . So we h ave the required total as 9 3 i n scale- l 0 notation. B ut 93 = 2 . 62 + 3 · 6 + 3 , which in scale-6 notation would be writte n as 233. Hence we would answer the que stion in word s as: "two three thre e . " Of course denary scale word s, such as twenty, h u ndred , etc . , cannot be used when working in a non-d e n ary scale notatio n .

SOLUTIONS TO TEASERS

59

4 7 . GRANDPA'S B I RTHDAY

Age s : Doug, x years ; grandfather, ( l Oy + z ) years . Then, x (x + 1 )/2 = lOy + z + 1 , and x = y + z. Thence, x 2 - x = 1 8y + 2 , so (2x - 1 ) 2 = 72y + 9. Say (2x - 1 ) = 3k , m aking k 2 = 8y + 1 , with y < 1 0 . Tabul ate : k 2 = 9 25 49 2x - 1 = 9 15 2 1 X = 5 8 11 3 6 Y = 1 z = 4 5 5 Grandfather 14 35 65 5 8 11 Doug Obviously the age s were 6 5 and 1 1 years. 48. VANDALISM

List the Lowest Co m m o n Multiples of the possible sets of correct fraction denominators : 4, 5, 7, 9-L . C . M . 1260 3, 5 , 7, 9-L . C . M . 3 1 5 3, 4 , 7 , 9-L . C . M . 252 3, 4, 5 , 9-L . C . M . 1 8 0 3, 4, 5, 7-L . C . M . 4 2 0 But there were o n l y 300 birds origi n ally, so the n u mber re m aining m ust have been 252 or 180. And the i ncorrect fraction must have been 1/5 or 117. But more than 1 0 0 birds escape d , so 1 8 0 must h ave re­ mained, the fraction 117 being incorrect. Then there re m a i ne d : 60 finches 45 budgies 36 can ari e s 20 parrots and 19 myn a h bird s and others. Now, the origi nal n u m ber of canaries was e q u al to three times the n u m ber of parrots re maini ng, so there h ad been 60 canarie s , of which 24 canaries escape d . 49. TRANSPORTATION

Say 3 groups A, B, C of 1 0 e ach, starting time zero. (1) Truck takes A a distance of x miles, t aki ng x/40 hours . (2) Me anwhile B and C walk x / l 0 miles, so B and C are 9x/ 1 0 miles behind A when truck drops group A . (3) Truck drive s back, picks u p B and transports t o j oin

60

CHALLENGING MATHEMATICAL TEASERS

group A that has continued on foot, and drives back agai n . Truck picks up group C and transports right to Tulla to arrive there at the same time as A and B . (4) S o truck m ake s 4 trips after droppi ng group A : to "close" 9x/ 1 0 miles at 34 miles per hour, 9x/340 hours ; to overtake 9x/ 1 0 miles at 36 m . p . h . , 9x/360 hours ; to "close" 9x/ 1 0 miles at 34 m . p . h . , 9x 340 ho urs ; to overtake 9x/ 1 0 miles at 36 m . p . h . , 9x/360 hours. Those 4 trips take total of 7x/68 hours. (5) B ut, while the truck does those 4 trips, group A walks fi nal (24 - x) miles at 4 m . p . h . , taking (24 - x )/4 hours. He nce 7x/68 = (24 - x )/4, whe nce x = 1 7 . So total t i m e for the complete operation is ( 1 7/40 + 7/4), i . e . , 87/40 hours : 2 hours, 1 0 % minutes. 50. J UST FOR FUN

N

= 10000x + y = y 2/2 - x , whence y 2 - 2y so (y - 1 )2 = 20002x + 1 .

=

20002x ,

Hence , (y - 1 ) == ± l (mod 274) == ± l (mod 73). Say (y - 1 ) = 274u + a = 73v + b , e ntailing u = 73k - 4a + 4b , say. T h e n (y - 1) = 20002k - 1 09 5a + 1 096b , where a = ± 1 , b = ± 1 . The nce (y - 1 ) == ± 1 or ± 2 1 9 1 (mod 20002). But (y - 1 ) < 9999, so (y - 1) = 1 or 2 1 9 1 . With (y - 1 ) = 1 , w e h ave x = 0, obviously unacceptable . He nce (y - 1 ) = 2 1 9 1 , making x = 240, y = 2 1 92 . The complete n u m b e r was 240 - 2 1 92. 51. GOOD SERVICE

Poole trains le ave at intervals of x m i n utes, Tulla trai ns at intervals of y minute s ; x and y being integers. Counting i n minutes from 6 : 00 a . m . , 1 8x = 1 7y + 5 , whence x = 1 7k + 5 , y = 1 8k + 5 . But, i n approximately 1 9 hours, between say 1 080 and 1 1 40 m i nutes fro m 6 : 00 a . m . , departures coincide only twice , so that xy > 2 5 , and also xy < 1 599 (i.e., 39 . 4 1 ).

SOLUTIONS TO TEASERS

61

S o w e must have k = 1 , making x = 2 2 , y = 23. He nce the fi rst coincide nce is at 1 2 : 3 6 p . m . (sh ortly after noon). The i ntervals being 22 and 23 minute s ; the next coinci­ dence occurs 506 mi nutes l ater, which is at 9 : 02 p . m .

52. B OTH HAN D S

S a y t i m e " i s " : x hours, y minute s , with h o u r h an d at t h e (60x + y )/ 1 2 mi nute division, and the m i n ute h and at the y m i nute s division. E arlier that day the time had been z hours, w m i n ute s , with h o u r h a n d a t t h e (60z + w )/ 1 2 min ute division, and the m i nute hand at the w minutes division. Then, (60z + w )/ 1 2 = y , and w = (60x + y )/ 1 2 - 1, whence 720z + 60x - 1 43y = 12. All of x, y, z, w are integers, so that e ntails y = 1 2k . The n, 60z + 5x - 1 43k = 1 , he nce k = 5 t - 2 . B ut Y < 60, s o k < 5 , hence w e require k = 3 , making y 36. The nce 5x = 430 - 60z , so x = 86 - 1 2z . But x < 1 2 , s o w e must have z = 7 , with x = 2 . W e had w = (60x + y )/ 1 2 - 1, s o w = 1 2 .

=

When J o e asked , the time was 2 : 36 p. m . ; when B e n checke d his watch e arlier t h a t day the t i m e had b e e n 7:12 a.m.

5 3 . LONG O D D S

O d d s of 1 1 to 2 agai nst i m p l i e s a chance of 2/ 1 3 . S a y Pat picke d up ( y + 1 ) card s , ( x + 1 ) being spad e s . ( x + l )x (x - 1 ) = 2 -, hence 1 3x (x 2 - 1 ) = 2y (y 2 - 1 ) . The n, ( y + 1 )y (y - 1 ) 1 3 B u t ( x + 1 ) < 1 4 , s o 2y (y 2 - 1 ) < 1 6 9 · 1 6 8 , y (y 2 - 1 ) < 1 4 1 96 . But 25(252 - 1 ) = 1 5600, so y < 25. Now, y or (y - 1 ) or (y + 1 ) must be a multiple of 1 3 , hence y = 1 2 or 1 3 or 1 4 .

62

CHALLENGING MATHEMATICAL TEASERS

Substituting for y, neither y = 12 nor y = 1 4 gives an integral value for x. But with y = 13, we h ave x = 7. So Pat picked u p 1 4 cards, including 8 spades. 54. TH E TAL KATIVE GUE ST

Remember the old chestnut in logic. It is true that "All m e n with re d hair are males," but surely not true th at "All males are m e n with red h air" ! List the various st ate ments : ( 1 ) Doug d e ntist if B e n banker. (2) Doug actor if Andy curate . (3) B e n actor O N L Y if Doug dentist. (4) Ben banker if Clem actor. (5) Doug curate U N L E S S Cle m dentist O R Ben banker. Ben is not the curate . If h e is the banker, Doug must be the dentist ( 1 ), which m ake s Ben the actor (3). If B e n is the actor, Doug must be the dentist (3). Then , re (2), Andy is not the c urate . So Andy would be the banker and Clem the curate : i m possible re (5). So Ben i s the d e ntist, with Doug the curate (5). Then, re (4), Clem is not the actor ; so Clem must be the banker and Andy the actor. Andy was the actor, Ben the dentist, Clem the banker, Doug the c urate . 55. URAN I U M TRIANG L E S

Say sides were 1 7, 28, x fee t and 17, 28, y feet. Any triangl e w i t h sides a , b, c has are a vs (s - a )(s - b )(s - c ) , where a + b + c = 2s . So, using the formula and squ aring, we get : (452 - X 2)(X 2 - 1 1 2) = (452 - y 2)(y2 - 1 1 2) . The n , setting X = X 2, Y = y 2, and sim plifying, (X - 1 073)2 = (Y - 1 0 73)2. But we re q u ire x and y to be different, hence we must h ave (X - 1 973) = - (Y - 1 0 73), whence X + Y = 2 1 46. So, x 2 + y 2 = 2 1 46 = 1 1 2 + 452 = 252 + 392. B u t (28, 1 7 , 1 1 ) and (28, 1 7, 45) cannot be sides of triangles. He nce the re s pective third sides were 2 5 and 39 feet.

SOLUTIONS TO TEASERS

63

56. MEN ON THE MOON

Say caches were set u p at points A, B, C distant fro m starting point, at a miles, ( a + b ) miles, and ( a + b + c ) m i l e s re s pective ly. We c a l l a unit of fuel a fuel-mile , and the crawler c arri e s 3 1 5 such units. (1) 3 round-trip runs to A, e ach time leaving (x - 2a ) units. l one-way run to A, arri ving with spare (x - a ) units. Then at A there will be available (4x - 7a ) units. (2) 2 round-trip run s to B, e ach ti m e leaving (x - 2b ) units. l one-way run to B, arriving with spare (x - b) units. Then at B there will be available (3x - 5b) units. Fuel expe nded, A to B, totals 5b units. So, 4x - 7a = 3x - 5b + 5b , whence x = 7a . (3) 1 round-trip run to C , le aving (x - 2c ) units. l one-way run to C , arri ving with spare (x - c) units. Then at C there will be available (2x - 3c ) units. Fuel expe nded, B to C , totals 3c units. So, 3x - 5b = 2x - 3c + 3c , whence x = 5b . (4) 1 final one-way run, using x units. Then 2x - 3c = x, whe nce x = 3c . But x = 3 1 5 , so a = 4 5 , b = 63, c = 1 0 5 . So caches were set up at 45, 108 and 213 miles from the starting point. All runs were made fully laden with fue l as shown in the detailed analysis. Total distance trave led by the crawler was 1 260 miles. 57. A FRIENDLY TE LLER

Check: ( l Ox + y ) dollars, ( 1 0z + w ) cents. Payment: ( l Ow + z ) dollars, ( l Oy + x ) cents. So 2000x + 200y + 20z + 2w = 1 000w + 1 00z + l Oy + x + 1 2 5 , whe nce 1 999x + 1 9 0y - 80z - 998w = 1 2 5 . Obviously, w m ust be twice or three t i m e s x , so x < 5 . Also, from t h e e q u ation, x m ust be odd. Say x = 3. The n 40z - 95y + 499w = 2936, e ntailing w = 9. But that leads to 1 9y - 8z = 3 1 1 , which is impossible since y < 1 0 .

64

CHALLENGING MATHEMATICAL TEASERS

S ay x = 1. Then 40z - 95y + 499w = 937, entailing w Thence, 1 9y - 8z = 1 1 2, so y = 8, Z = 5.

=

3.

So the check was for $ 1 8. 5 3 , the payment $35 . 8 1 . 5 8 . MANY A M I C KLE Say there were (n + 1 ) amounts listed, and that e ach ch arity would receive m ¢ . T h e n , 24 + 1 2 2 ( 1 + 2 + 4 + . . . + 2 n - l ) - 5 2 0 0 = 1 1m , s o 1 2 2 · 2 n = 1 1m + 5 2 9 8 , whe nce 6 1 · 2 n == 9(mod 1 1 ), which e ntails 2 n == - 4(mod 1 1 ). Now, all to (mod 1 1 ) we have 22 == 4, and 25 == - 1 , so 2 7 == - 4, and 2 7 == 2 1 7 == 22 7 , etc. He nce n = 7, or 1 7, or 27, etc. With n = 7, we have m = 938, and with n = 27, m > 1 0 9 • Obviously e ach ch arity would re ceive more than $9.38, so we must have n = 1 7, with m = 1 453226. So e ach ch arity would rece ive $ 1 4,532.26. 59. A TALE OF WOE

T ake the second day as No. 1 , sales the first day being x ¢ . T h e n on succe s sive days s a l e s i n cents were : N o . 0 : x , No. 1 : (2x + 3)/3, No. 2 : (4x + 24)/9, No. 3 : (8x + 1 29)/27, No. 4: ( 1 6x + 582)/8 1 , etc. So sales for day number n were (2 nx + 3U n )/3 n , where U I = 1 , U = 8, U = 43, U = 1 94 , and U = 2 U n3 n n - I + 3 ln . 2 4 S a y U n = n · 3 nA + n · 2 nB + 3 n C + 2 nD + E . Obviously U 0 = O. Then, su bstituting our known numerical values of U n for n = 0 , 1 , 2 , 3, 4 we h ave a system of 5 linear equations. Solving t h at syste m we fin d : A = 1 , B = 0, C = - 2 , D = 2 , E = O. Hence U n = 3 n ·n - 2 · 3 n + 2 · 2 n = 3 n (n - 2) + 2 n + l . So sales for d a y n u m ber n were : 2 n (x + 6)/3 n + 3n - 6, which e q u aled 533 cents . H e n c e , 2 n (x + 6)/3 n + 3n = 539. Then (x + 6) must be divisible by 3 n , so (x + 6) where k is a n integer, whence 2 nk = 5 3 9 - 3n .

3 nk ,

SOLUTIONS TO TEASERS

65

From the wording, n > 3. A l s o , since 2 10 = 1 0 2 4 , n < 1 0 . A n d ( 5 3 9 - 3n) m u s t be e v e n , so n m u s t be o d d . Testing for n = 5 , 7, and 9 w e h ave integral k o n l y with n = 9, k = 1. Hence x = 19677. The first day sales amounted to $ 1 96 . 77 , and they were in business for 1 0 working days .

60. QUITE A F AMIL Y

Say age s were : boys (x - 2), x , (x + 2) ye ars ; girls (y - 1 ) , (y + 1 ) years. Then x 2 - 2y 2 = - 2 . Say x = 2z , whence y 2 - 2Z 2 = 1 . Tabul ate solution s : y = 1 3 1 7 etc. Z = 0 2 1 2 etc. x = 0 4 24 etc . , x, y, z a l l obe ying relation U n + = 6 U n + 1 - U n. 2 No further solution s could yield acceptable age s , and John said the age s we re all differe nt. Hence we must have x = 24, y = 1 7 , m aking the age s : boys 22, 24, 26 years, and girl s 16, 1 8 ye ars .

6 1 . NUMBERS, N U M B E RS

Say the two " h alve s " were X and Y, e ach < 1 000. Then, X2 + P + 1 = 1 0 0 0X + Y, so (2X - 1 000)2 + (2Y - 1 )2 = 999997 = 1 32 1 · 7 5 7 . 1 32 1 · 7 5 7 = (36 2 + 5 2)(26 2 + 9 2) = (36 ·26 ± 9 · 5) 2 + (36 ·9 =+= 2 6 · 5) 2 = (891 2 + 454 2) or (98 1 2 + 194 2) . Tabulate : 2X - 1000 2Y - 1 X Y

= - 4 54 = 89 1 = 2 73 = 446

+ 454 891 727 446

- 1 94 981 403 491

+ 1 94 981 597 491

So there were 3 other 6-digit nu mbers that met the re ­ quire m e nts : 273446, 59749 1 and 727446.

66

CHALLENGING MATHEMATICAL TEASERS 62. THE YEARS THAT COUNT

Age s : Sam x ; boys, y and 9; wife , z years. x 2 + y 2 + 92 The n , = z, and ( x + y + 9) is a factor o f x + y + 9 (x 2 + y 2 + 9 2 ). That entails ge neral solution : x = a(a + b + c ) Y = b(a + b + c ) 9 = c (a + b + c ) with z = a 2 + b 2 + c 2 a, b, c being i ntegers. From c (a + b + c ) we have c = 9, or 3, or 1 . With c = 9 o r with c = 3, w e would h ave negative values for a and b . So, c = 1 , whe nce a + b = 8. The n , taking a > b , and since obviously y > 9 and z > (y + 10) say, we must h ave : a = 6, b = 2, x = 54, y = 1 8 , z = 4 1 . S a m w a s 5 4 ye ars old, h i s wife 4 1 , t h e boys 1 8 a n d 9 . 63. A FUNNY FRACTION

Say the amount was x doll ars and y cents. T h e n 1 0 ny + x = (5 00x + 5y )/8, where x has n digits. He nce y (8 · 1 0 n - 5) = 492x . Now set x = 5w , y = 4z , whence z ( 1 6 · l O n - 1 - 1 ) = 1 23w . B u t ( 1 6 · l O n - 1 - 1 ) i s a multiple of 3 for all n , so ( 1 6 · l O n- 1 - 1 ) = 3k , whence zk = 4 1w . T h e n , since y < 1 0 0 , w e h ave z < 2 5 , s o k must b e a m ultiple of 4 1 , whe nce 1 6 · l O n - 1 == l(mod 4 1 ). Now say 1 6 · l O n - 1 = 4 l t + 1 , which e ntails I O n- I = 4 1 m + 1 8 . So I O n- I == 1 8(mod 4 1 ) . B y quick tri al w e fi nd t h e m i n i m al solution with n = 3. The nce, fro m our original e q u ation, 65y = 4x , with general solution x = 65r, y = 4r , say. The m i n i m al value for a 3-digit x e ntails r = 2 , making x = 130, y = 8. The amount was $ 1 30 . 0 8 . 6 4 . TWO MAG I C SQUARES

Say a M agic Squ are has n u m bers (x + 1 ) to (x + r 2 ), where x + r2 = 5 4 .

SOLUTIONS TO TEASERS

67

Then it h a s r2 n u m bers from (55 - r2) to 54 inclusive, m aking a gros s total of r2( 1 09 - r2)/2, a magic total of r( 109 - r2)/2. Say our two M agic Squ are s have m2 and n 2 n u m bers re­ spectively, with m > n. Then m ( 1 0 9 - m2)/2 = n ( 1 0 9 - n2)/2, whence m3 - n3 = 1 09(m - n ) . Now, m i= n, so we m ay divide through by ( m - n), giving m 2 + mn + n 2 = 1 09, whe nce (2m + n)2 + 3n2 = 1 0 9 · 4 = ( 12 + 3 . 62)(12 + 3· 12), with integral solutions m = 7, n = 5 , or m = 5 , n = 7. But m > n , so we must have m = 7, n = 5 . He nce t h e respective M agic Squares m u s t h ave n u mbers from 6 to 54, and 30 to 54, respectively. 65. THE JOKER'S WI LD

They starte d wit h : Jack x cents, Bob y cents. After n wins by Jack they had: Jack: x + y(3 n - 2 n )/3 n , Bob: 2Ry/3 n . After a further n wins by B o b : Jack had 2 n [x + y (3 n - 2 n )/3 n ]! 3 n . But that equaled (x + y )/4. Hence, 3 n ( 4 · 2 n - 3 n )x = [32 n - 4 · 2 n (3 n - 2 n )]y , with solution : x = (32 n - 4 · 2 n · 3 n + 4 ' 22 n )k , y = 3 n (4 ' 2 n - 3 n )k . So, (x + y )/4 = 22 nk , where k is an integer. Now Jack lost, so x > (x + y )/4. Say 3 n = a, and 2 n = b. The n , a2 + 3b 2 > 4a b , hence a > 3b . That e ntails 3 n > 3 · 2 n , so n > 2. But we must have 4 · 2 n > 3 n , so n < 4. Hence, n = 3, making x = 1 2 1k, y = 1 3 5k, (x + y ) = 2 5 6k . J ack ended u p with 64k s o h e lost 5 7k cents. B ut that was " almost exactly $4," so k = 7. Then x = 847, y = 945. J ack starte d with $8.47, Bob with $9.45. 66. EASY COME EASY GO

Say y pairs at $ 1 .35, x n u m be r of cents on the invoice . Then, 1 3 5y = 100(x 2 + 3) + x , whence 1 00x 2 + x + 300 135y , and 1 3 5 = 2 7 · 5 .

=

68

CHALLENGING MATHEMATICAL TEASERS

Dividing ( 1 00x 2 + x + 300) by 1 0 0 we h ave (x 2 + x / 1 0 0 + 3). 27·37 + 1 1 == l O(mod 27). = 1 0 , so Now, 100 100 H e nce, x 2 + l Ox + 3 == O(mod 27), whe nce (x + 5 ) == ± 7(mod 27). So (x + 5 ) = 27k ± 7, say. B ut, (x 2 + 3) < 500, so x < 23. Obviously x *- 2. H e nce we must have (x + 5 ) = 20, x = 15 . So the total amount w a s $228. 1 5 , for 169 pairs . 67. A BUG IN A B ARN

It is well known t h at the angle of incidence equals the angle of reflection, the p ath of such a ray being the sh orte st possible. He nce the route of the roach would corre s pond to that of a ray of light reflected by the front and back walls. So, the diagrams depict the actual and virtual situ ations, the angles at D and E all being e q u al . VI RTUAL

A CT U A L

90

E

So we have (B C )2 = 5 6 2 + 902, m aking B C T h e shortest possible route was 106 feet.

=

106.

68. WRONG N U M B ER

S ay J o e ' s n u m be r was ( 1 00y + l Ox + z ), where x and z are si ngle-digit n u m bers, and y is an i nteger less than 1 00.

SOLUTIONS TO TEASERS

69

Then 1 000x + l Oy + z = 300y + 30x + 3z, whence 5(97x 29y ) = z . Then we must h ave z = 5 , making 97x - 29y = 1 . That e q u ation h a s ge neral integral soluti o n : x = 29k + 3 y = 97k + 1 0 Within perm issible limits, therefore , x = 3, y = 1 0 . S o , Joe's n u mber w a s 1035. -

69. A T TH E SHOW

Say X boys, Y girl s. T h e n the total n u m be r of different arrange m e nts i n one row i s l (X + Y). The n the total n u m be r of differe nt arrange ments with a girl at e ach end would be Y(Y - 1 ) I (X + Y - 2), so chances of a girl at e ach end will be : Y(Y - 1 ) Y(Y - 1 ) I (X + Y - 2) . , I.e., I (X + Y) (X + Y)(X + Y - 1 ) " But "2 to 1 o d d s against" is one ch ance in thre e , 1 Y(Y 1) so = (X + Y)(X + Y 1) 3" ' whence 2Y2 - 2XY - X2 = 2Y - X. Setting k = 2Y - X, this becom e s k2 - 2k = 3X2, whe nce (k - 1 )2 - 3X2 = 1 . This Pe lli an e q u ation h a s succe ssive integral solutions that m ay be tabulated as: (k - 1 ) = 1 2 7 26 97 etc., X = 0 1 4 1 5 56 etc . , -

-

with Y = 1 2 6 2 1 77 etc. But we re quire X > 4, re 5 B arton boys, and (X + Y) < 1 0 0 . So w e m ust h ave X = 1 5 , Y = 2 1 . He nce there were 3 6 in the group: 2 1 girls a n d 1 5 boys.

70. THE PE NALTY C LAUSE

Say x plants , and y minutes pen alty time. Then, 37x - y(y + 1 )/2 = 700, whence (2y + 1)2 = 296x - 5599, and 296 = 3 7 · 8 . Hence (2y + 1 ) 2 = - 5 599(mod 8) = - 55 99(mod 3 7) so (2y + 1 ) = ± l (mod 2) = ± 5(mod 37).

70

CHALLENGING MATHEMATICAL TEASERS

Thence, (2y + 1) == ± 5(mod 74). B ut we h ave "nearly 90 of them in all , " so x < 90, whence we must h ave y (y + 1) < 5260, entailing y < 73, and (2y + 1) < 1 4 7 . Pos sible v a l u e s may be tabulated : 2y + 1 = 69 79 143 71 m aking y = 34 39 88 with x = 3 5 40 Then, for " n e arly 90," x = 88, with y = 7 1 . There were 8 8 rose stand ard s in all. 7 1 . CONSEC UTIVE SQUARE S

Say N = 10000y + 6789, y being a 3-digit integer. Then 10000y + 6789 = (x - 1)2 + x 2 + (x + 1)2 = 3x 2 - 2, say. Dividing through by 3, we see t h at y = 3z - 1 , say. So x 2 = 10000z - 1 0 7 1 , whence x 2 == - 1 07 1 (mod 1 0000), giving x2 == l(mod 16) == 1 79(mod 625). Say f(x) = x 2 - 179 == O(mod 5), making x = 5a ± 2, say. Then f(x ) = 25a2 ± 20a - 175 == O(mod 25), hence 5a 2 ± 4a - 3 5 == O(mod 5), so a = 5b , say, m aking x = 2 5b ± 2. Then f(x ) = 252b 2 ± 1 00b - 175 == O(mod 1 2 5), hence 25b2 ± 4b - 7 == O(mod 5), so b = 5c ± 3, m aking x = 1 2 5c ± 77. Then f(x ) = 12 52c 2 ± 1 2 5 · 1 54c + 5750 == O(mod 625), hence 1 2 5c 2 ± 1 5 4c + 46 == O(mod 5), so c = 5d ± 1 , m aking x = 625d ± 202. So we h ave x == ± 202(mod 625) == ± l(mod 8), which entails x == ± 423 or ± 1 673(mod 5000). Hence , since we require 3x 2 < 9999999, x = 423 or 1 673. B ut with x = 423 we h ave N = 536789, a 6-digit n u m ber. Hence we must h ave x = 1673, m aking N = 8396789. 72. THI N K OF A NUMBER

Say the overall are a of the squ are p atio was x 2 squ are feet, the internal rectangular lawn h aving an are a of 414 squ are fe et. Then x 2 - 4 1 4 = 7y , whence x 2 == 1(mod 7). So x == ± l(mod 7). Now, x 2 - 414 < 414, whence x < 29.

SOLUTIONS TO TEASERS

71

Al so, the longer s i d e of the i nternal lawn i s 23 feet, so x > 23. Hence we must h ave x = 27, m aking the p aved are a 3 1 5 . S o Fran 's key n u mber w a s 4 5 . 7 3 . A DEAL I N STAMPS

Say, x stamps at x cents, y stamps at 5y cents . The n , x2 - 5y 2 = 3 1 9 , and x 2 + 5y 2 = about 3200. Now, 319 = 2 9 · 1 1 , so the equ ation can be solved an alyti­ cally, there being two families of integral p airs of (x , y) v alu e s . But here it seems th at i nspired tri al i s j u stifi e d . Adding t h e two equ ation s , 2X 2 = about 3 5 1 9 , so we must h ave x = about 42. Obviously, from the first e q u ation, x 2 ends with 4 or 9, so we need test only x = 38, 42, 43, and 4 7 : x = 38 42 43 47 X 2 = 1444 1 764 1849 2209 m aking 5y 2 = 1 1 2 5 1445 1530 1890 y2 = 225 289 306 378 i ntegral y = 15 17 x 2 + 5y 2 = 2569 3209 We may rej ect the first case, since all requirements are obviously met by the results in the second. Hence , he bought 42 stamps at 42¢ average , and 1 7 at 85¢ average , for a total of $32.09. 74. A SERIAL NUMBER

Say the seri al n u mber N = ( 1 000000 1x + lOy), where y < 106• Then N = ( l 1x )5, so y i s a m ultiple of 1 1 and of x . Say y = 1 1xz , whence 1464 1x 4 = 1 0z + 90909 1 . Now since X 4 > 90909 1/1464 1 , x > 2 . 1 1xz < 1000000, s o x z < 90909, a n d x 2: 3 , s o z < 30303, whence 14641x 4 < (303030 + 90909 1 ) . Hence X 4 < 83, so x < 4 . B u t , x 2: 3. Hence x = 3, making z = 2 7 6 8 3 , y = 913539. The serial n u mber was 3 9 1 3 5393. 75. AN EVENING OUT

Stan started with : x dollars in bills, 2y cents in coi n s , m aking the total ( 1 00x + 2y ) cents .

72

CHALLENGING MATHEMATICAL TEASERS

On arrival back home he h ad left ( 1 00x + 2y)/3 cents, so Fiona left h i m [(100x + 2y)/3 - 50] cents . Say this was y dollars and x cents , totaling (x + 10Oy) ¢ . Then , ( 1 00x + 2 y - 1 50)/3 = x + 1 0Oy , whence 97x - 298y = 1 50, with general solution : x = 298k + 1 06, y = 97k + 34. Then, 1 00x + 2y = 29994k + 1 0668. Two-third s of that i s 1 9996k + 7 1 1 2 , which was less than 1 0000. So here we require k = 0, making x = 1 06 , 2y = 68. Stan st arted with $ 1 06.68. 76. A MAITE R OF AGE S

Age s : Ted x years, Robin y , Pat z , all integral, x > y . Thence, x 2 - y 2 = Z 3; and x/y = y/z , whence y2 = xz . Substituting for y2 we h ave x 2 - XZ = Z 3, so x (x - z) = Z 3. That equ ation i s fu lly satisfied by: x - z = mz , and x = Z 2/m• where m i s any rational number. Thence , z = Z 2/m - mz , so z = m (m + 1). Su bstituting for z, we h ave x = m (m + 1)2, whence y 2 = m2(m + 1)3, which entails y =m v(m + 1)3. Now, y i s an integer, so here we see m must be an integer. Then , (m + 1) must be a square, so s ay m + 1 = n2• The nce, m = n2 - 1 , and substituting we h ave : x = n4(n2 - 1), y = n3(n2 - 1), z = n2(n2 - 1). Obviously, n > 1 . Also, for possible h u man age s , n < 3, so n = 2, whence age s were : Ted 48 ye ars , Robin 24 ye ars, Pat 1 2 years . 77. DRO PPI NG IN ON LEN

I

A L-----

-+�------�x�------� B p

��____

__

SOLUTIONS TO TEASERS

73

= PB = P C , so A, B , and C all lie on a circle with center at P. AB i s straight, so must be the di ameter of th at circle , m aking the angle at C a right angle .

PA

Then, (2X)2 = (3Z )2 + y2. Also, x + Z = 2 3 , whence 5z 2 + 1 84z + y2 = 2 1 1 6 , hence (5z + 92)2 + 5y 2 = 4 . 692. 692 = (82 + 5 · P)(82 + 5 · P) = (82 + 5 · P)(72 + 5 . 22) = (72 + 5 . 22)(72 + 5 . 22) = 592 + 5 . 162 = 462 + 5 . 232 = 662 + 5 . 92 = 292 + 5 . 282. Thence, 5z + 92 = 1 1 8 or 92 or 132 or 58 8 0 m aking z = 18 with y = 46 So it is 18 miles from Brent to Crowe as the crow flie s . -

-

7 8 . CARDS ON THE TAB LE

Say tabletop in squ are 1/4 -inches was (x + 4) by x . E ach card h ad are a 126 squ are lit -inc h e s . He had an e v e n number of surplu s card s , so we m ay say: (52 - 2y ) c ard s occupied 2 52(26 - y ) squ are lit -inch e s . B ut he h ad 180 squ are %-inches o f wasted spac e . S o , x (x + 4) = 252(26 - y ) + 1 8 0 , whence ( x + 2)2 = 6 7 3 6 - 2 52y . Thence (x + 2)2 == 6736 (mod 252), so (x + 2)2 == 4(mod 9) == 2(mod 7) and i s even. Thence, (x + 2) == ± 1 1 or ± 2 5(mod 63) and is eve n . B u t (x + 2)2 < 6736, so ( x + 2) < 83 . Also y < 1 4 , s o (x + 2)2 > 3208, hence ( x + 2) > 5 6 . S o w e must h ave ( x + 2 ) = 7 4 , m aking x = 72. Thence, y = 5 . T h e tabletop w a s 18 i n c h e s by 1 9 inches.

79. A BUS RI DE

Starting with x passengers , the n u mbers c arried for e ach of the 5 successive stage s were : x, (2x + 1 5)/3 , (4x + 48)/ 9, (2x + 60)/9, (x + 39)/9. Total nu mber of different passengers was (x + 1 2 ) . Total n u mber of passenger-stage u nits was ( 2 2 x + 1 92)/9 , but th at was an integer, so s ay x = 9y + 6 . Thence tot al different passengers was ( 9y + 1 8), total

74

CHALLENGING MATHEMATICAL TEASERS

passenger-stage units was (22y + 36), total fares in dollars (9y + 18). Say, z adult passenger-stage units , (22y + 36 - z ) child passenger-stage units . Then, total fares would be z/2 + (22y + 36 - z )/4 dollars. Thence, (z + 22y + 36)/4 = 9y + 18, so z = 14y + 36. But one-fifth of all the different passengers were chil­ dren, so (9y + 18) i s a multiple of 5 , m aking y = ( 5w + 3), say. Thence, z = 70w + 78. At the start there were x passengers, and x = 9y + 6, so the bus started with (45w + 33), "fewer than 60" pas­ sengers. So w = 0, m aking x = 33, with y = 3, z = 78. Then there were 4 5 passengers, one-fifth being children. So there were 36 adult passengers . 80. YOU'RE S URE IF IN DOUBT One child must h ave been less th an 4 ye ars old. Tabulate all factors of 96 i n sets of three : 24 16 12 16 12 8 8 6 4 6 8 3 4 6 4 4 1 1 1 2 2 2 3 4 Totals 29

23

21

21

18

16

15

14

Now, Ted obviously knew the n u mber of the house . The fact th at h e was still i n d oubt shows that the age s must h ave totaled 2 1 . B ut, if the ages h a d been " 1 6, 3, 2 " there would h ave been only one child 4 years previously. Hence, the ages must h ave been 12, 8, and 1 years. 8 1 . C RE E PY C RAWLI E S Say x children, y slugs . I ndic ate individual children as : A got a slugs, B got b slugs, C got c slugs, etc . Then A received l Oa - 3(y - a ) = 13a - 3y , B received ( 13b - 3y), and so on.

SOLUTIONS TO TEASERS

75

Total net amount received by the childre n : 13(a + b + c + . . . ) - 3xy , where y = (a + b + c + . . . ) So, 13y - 3xy = 95x , whence ( 1 3 - 3x)(95 + 3y) = 13 · 95 Tabulate for factors : [NOT E : (13 - 3x) < 8] 13 - 3x = 5 1 95 + 3y = 247 1 2 3 5 x = 4 y = 380 So, 380 slugs were collected by 4 children. A possible bre akdown of the totals would be: A got 92 : received net $9.20- $8.64, i.e., $0.56 B got 94 : received net $9.40-$8.58, i.e., $0.82 C got 96: received net $9.60-$ 8 . 5 2 , i .e . , $ 1 .08 D got 98 : received net $9.80-$8.46, i . e . , $ 1 .34. -

-

82. FAM I LY NUMB E RS Say Doug's special divisor was x, the rem ainder being y in e ac h case. ax + y = 9638 Then say: b x + y = 8739 cx + y = 2 5 9 1 Subtracting: x(a - b) = 899 = 2 9 · 3 1 x ( b - c ) = 6 1 4 8 = 4 · 2 9 · 53 The common factor i s 29, so x = 29. Doug divided e ach number by 29, with remainder 1 0 in e ach cas e . 83. TOO MANY GIRLS

Say there were x girl s, e ach scoring y/2 points , and 2 boys. Then there were (x + l)(x + 2)/2 games i n all, and there were (x + l)(x + 2)/2 points in all. But the total nu mber of points was xy/2 + 9. So (x + l)(x + 2) = xy + 18, whence (2x - y)2 + 12x = y2 + 64. Say (2x - y) = X, m aking (X + 3)2 - (y - 3)2 8 16 Thenc e : [(X + 3) + (y - 3)]/2 = 4 2 1 with [ (X + 3 ) - (y - 3)]/2 = 4 m aking (X + 3) and

=

(y - 3)

=

8 0

10 6

17 15

4 3

8 9

16 18

-------

whence with

x y

=

=

=

64.

76

CHALLENGING MATHEMATICAL TEASERS

B ut "the girl s all did better than e ither of u s , " and the two boys totaled 9 points in all. Hence y/2 > 9/2 , y > 9 . So we must h ave x = 1 6 , with y = 1 8 . H e n c e there werf> 1 6 girl s, e ach scoring 9 points, (i. e . , e ach girl played 2 games agai nst boys, 15 agai nst girls). 84. WHAT? NO ZOBBLIES? Say 5x at 97¢ for 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97x ¢, 3y at 67 ¢ for 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67y ¢ , ( 1 0 0 - 5 x - 3y ) at 25¢ . . . . . . . . . . 2 5 ( 1 0 0 - 5 x - 3y ) ¢ . T h e n 2 5 ( 1 0 0 - 5 x - 3y ) + 97x + 67y = 2000, whence 7x + 2y = 125, with ge neral integral solutions : x = 2k + 1 , y = 59 - 7k . T h at entails (5x + 3y) = 1 82 - 1 1k < 1 0 0 , so here we must h ave k > 7. But, re y , k < 9. So k = 8, m aking x = 1 7 , y = 3, 5x = 8 5 , 3y = 9. Hence, 6 zobblies were sold at 2 5¢ each. 85. THE CE NSUS Say (3X )2/3 c ats , i . e . , 3x 2. Then, if a qu arter were slai n , 9x 2/4 would re main. So, 9x 2/4 = y 3 , say, so (3x/2)2 = y 3 . Thence, y = k 2, say. B ut y must be a multiple of 3 , hence k = 3t , say. Thence, (3x/2) = 2 7t 3 , so X = 1 8t 3 , Y = 9t 2. For m i n i m al n u mber of c ats we must h ave t = 1 , with x = 18. Hence there were a t least 9 7 2 cat s . 86. A W H I Z K I D Say 7 x questions, (7x - y ) answered correctly. Then 7(7x - y ) - y (y + 1 )/2 = 1 68 , x and y being integers . Thence y (y + 1 5 ) = 98x - 336, whence (2y + 15)2 = 392x - 1 1 19, and 392 = 49, 8, so (2y + 1 5)2 == 8(mod 49) == 400(mod 49) and is odd. (2y + 15) == ± 2 9(mod 98). Hence The boy an swered more th an three-qu arters correctly, so we re quire 7x > 4y .

SOLUTIONS TO TEASERS

77

With (2y + 1 5) = 29, we h ave y = 7, x = 5, which is acceptabl e . With (2y + 15) = 69, we h ave y = 2 7 , x = 1 5 , n o t acceptable. With (2y + 1 5) = 127, and in all gre ater c a s e s , the x/y ratio would be acceptabl e , but the num bers of questions would be quite unrealistic. Hence we h ave x = 5 , with y = 7. There were 35 questions, and John answered 28 correctly. 87. TIC KETS I N THE SWE E P Say Susan h ad n u m bers ( x - 1 ) , x , a n d ( x + 1 ) , Jack h ad number y , Jill nu mber z . The n , (x - 1 )2 + y 2 = (x + 1)2, whence y2 = 4x . So, x = k2 and y = 2k , where k is an integer. We also h ave X2 + Z 2 = (x + 1 )2, whence Z 2 = 2x + 1 . (A ) Hence , Z 2 = 2k2 + 1 , making Z 2 - 2k2 = 1 . We h ave x < 1000, so k < 32. Within those limits e q u ation (A ) h as only two non-z ero integral solutions . With z = 3, k = 2 , we h ave x = y = 4, which i s unacceptable because all ticket num bers were different. So z = 17, k = 12, m aking x = 144, y = 24. Susan had nu mbers 143, 144, 145; Jack 24, Jill 1 7 . 88. PROGRE SS Say the odd house nu mbers ran from 1 to (2m - 1), and the numbers of the houses purchased ran from (2n + 1) to 43, all being odd . S u m 1 to (2n - 1) Sum 4 5 to (2m - 1) Su m 1 to (2m - 1) So , n2 + m2 - 484 = m2/2, whence m2 + 2n2 = 968. Say m = 2M, n = 2N, making M2 + 2N2 = 242, with solution M = 12, N = 7, wh ence m = 24, n = 14. So the odd numbers ran from 1 to 47 inclu sive . The houses th at were bought ran from 29 to 43, all odd.

78

CHALLENGING MATHEMATICAL TEASERS 89. THRE E TI M E S

S ay N = 100x + y , where x h a s n digits and y < 100. Then, 1 0"11 + x = 300x + 3y , whence ( l o n 3)y = 299x . 299 = 2 3 · 13, s o : ( I o n 3) == O(mod 13) or O(mod 23) or O(mod 299). -

-

Say I O n == 3(mod 13). We h ave 1 0 4 == 3(mod 13), so for minimal v alues n = 4 . Thence, 299x = 9997y , whence x = 769k, y = 23k , say. But x must h ave 4 digits, and y must be odd. H e nce k = 3 , m aking x = 2307, y = 69. N = 230769. Say IO n == 3(mod 23). All to (mod 23) we h ave 1 02 == 8, 1 03 5 , 105 == - 4 , 1 06 == 6, etc. == 1 1 , 104 == Hence this case would entail N gre ater th an 230769. -

Say lOR == 3(mod 299). All to (mod 299) we h ave 1 0 3 == 1 03, 104 == 133, 105 == 134, 1 06 == 144, etc. Hence this case would entail N gre ater th an 230769. So Bill's number was 230769. 90. A TOUC HING TALE

Referri ng to the diagram, for m i n i m al (AB + B C ) we must h ave angles ABE and CBD e q u a l , so E B /B D 48/168 = 2/7 . =

SOLUTIONS TO TEASERS

79

So say EB = 2x , B D = 7x , m aking AB = 2 VX2 + 576, B C = 7 VX2 + 576, AC = 3 V9x2 + 1600, whe nce AB + B C = 9 VX2 + 576 = 3 V9x 2 + 1600 + 48. V9x 2 + 5 184 V9x 2 + 1600 16, (1 ) So, (9x 2 + 5 1 84) But, (9x 2 + 1 6 0 0 ) = 3584. (2 ) Divide (2 ) by (1 ) : V9x 2 + 5 1 84 + V9x 2 + 1600 = 224. (3 ) = 240. Adding (1 ) to (3 ) : 2V9x 2 + 5 1 84 So V9x 2 + 5 1 84 = 120, whence x = 32. He nce the squ are fi eld had sides 288 feet, are a 82944 square feet. r--=------,-,�

91. FUN F O R SOME

Age s : J ack y, M arth a x. Then, 3y 2 - 2x 2 = 1 00x + y , whence (6y - 1)2 - 6(2x + 5 0)2 = - 1 4999 = - 2 83 ' 53, so (6y - 1 )2 - 6(2x + 50)2 = ( 1 72 - 6 . 12)(12 - 6 . 32) = ( 1 7 ± 18)2 - 6(5 1 ± 1)2 = 12 - 6 . 502 or 352 - 6 . 522. Re ( 1 2 - 6 . 502) :

6y - 1 = ± (m ± 30On) 2x + 5 0 = 50m ± n where m2 - 6n2 = 1 . Tabulating: m = 1 5 5 etc. 2 etc. n = 0 2 y = - - 101 etc. x = - - 101 etc. B ut we require x < 100 (i . e . , 2-digit integer). Re (352 - 6 , 522) :

6y - 1 = ±(35m ± 3 1 2n) 2x + 5 0 = 52m ± 3 5n where m2 - 6n2 = 1 . Tabulating: m = 1 5 5 4 9 etc. n = 0 2 2 2 0 etc. y = 6 75 - etc. x = 1 70 - etc. Obviously the ages were : Jack 75, M arth a 70 years . 92. HE B ROKE THE B A N K

Say, [x (x + y ) - 2 8 8 ] qu arters, x dimes, y nickels, x (x + y ) pennie s . T h e only even pri me number i s 2 .

80

CHALLENGING MATHEMATICAL TEASERS

If

X = 2 , there would be (2y - 284) qu arters. For that to be prime, we woul d require y = 143, composite . I f y = 2, there would be [(x + 1)2 - 289] qu arters, i . e . , (x + y 17)(x + y + 1 7 ) , composite . -

Also, x (x + y) - 288 must be prime. Since x and y are both odd primes, [x (x + y) - 288] must be even. Hence, x (x + y) - 288 = 2, x (x + y) = 290. Then, 290 = 2 · 5 · 2 9 , and (x + y) is even. Tabulating for factors : x + y = 290 58 x = 1 5 with y = 289 53 m aking But both x and y must be prime, so we h ave x = 5, y = 53. There were 2 qu arters, 5 dimes, 53 nickels, 290 pennie s, a total of $6.55.

93. DO IT YOURSE LF

We h ave, s ay, X 2 + y2 = Z2, and w2 + 232 = 2z 2, where x, y, z are integers, and Z 2 is "getting on for 1500." Then, w 2 - 2Z 2 = - 529 with two distinct basic solution s : w = 23, z = 23, AND w= 7, z = 17. So al l solutions will stem from two "famili e s . " Re (w ; z ) = (23 ; 2 3 ) we h ave : w = ± 23(m ± 2n), Z = 23(m ± n), where m2 - 2n2 = 1. B ut, Z 2 < 1 500, hence z < 39, so (m ± n) < 2 . That restricts us to w = 23, Z = 23, which is not acceptable because it woul d le ad to x 2 + y2 = 529 which can h ave no integral solution ( i . e . , 23 being of form [4r - 1]). Re (w ; z ) = ( 7 ; 1 7) we h ave : w = ± ( 7m ± 34n), z = ( 1 7m ± 7n), where m2 - 2n2 = 1. With m = 1 , n = 0, we h ave w = 7 , Z = 1 7 . But we require Z2 to be "getting on for 1 5 0 0 . " W i t h the next larger p a i r of (m, n ) values, m = 3, n = 2, we h ave w = 47, Z = 37, and 3 72 = 1369. Then, since 1369 = 122 + 352, Joe ' s new tables would h ave 144 and 1 2 2 5 tiles respectively, a total of 1369 tile s .

81

SOLUTIONS TO TEASERS 9 4 . A B UG F O R THE BIRDS

S\Z X51

I,

37

A.

I

_ _

_

J

a

a

Are a of tri angle with sides x , y , z i s : vs (s - x )(s - y )(s - z) where x + y + z = 2s . Hence are a of tri angle A i s V(882 - X 2)(X2 - 196)/4 =xh/2 , and are a of tri angle B i s V( 1 1 62 - X 2)(X 2 - 1 96/4 = xa/2. So (882 - X2)(X 2 - 1 96) = 4x 2h2 and ( 1 1 62 - X 2)(X 2 - 1 96) = 4x 2a2. B ut a2 + h2 = 512 = 260 1 , hence 4x 2(a2 + h2) = 1 0404x 2. Thence , (x 2 - 1 96)(882 - x 2 + 1 1 62 - x 2) = 1 0404x 2, whence X 4 - 5 594x 2 + 2077600 = 0, so x 2 = 2797 ± 2397. With X2 = 5 1 94 , the bug would h ave to be right on the feeder tray wh ich, incidentally, would be unduly large . So we h ave x2 = 400, m aking x = 20. The feeder tray had 20-inch si d e s . 95. A LUCKY NUMBER

Say : N = 1 05a + 1 0 4b + 1 0 3c + 102d + 1 0e + f and kN = l OY + 10 4e + 1 0 3d + 1 02c + l Ob + a. The n , 105(ak - j) + 1 04(bk - e ) + 1 0 3 (c k - d) + 1 02(dk - c) + 1 0(ek - b ) + fk = a, whence fk = l Ot + a, where t < 9, with if - k ) < ak � f, and ak < 1 0 , f < 1 0 . Obviously, if k i s e v e n , a will be e ve n , also k > 1 , s o a < 5. Hence , a = 4, 3 , 2, or 1 . All re quire ments are met o n l y with : a = 2, k = 4 , fk 32, f = 8 O R a = 1, k = 9, fk = 8 1 , f = 9. =

82

CHALLENGING MATHEMATICAL TEASERS

Say, a = 1 , k = 9, f = 9. Then 104(9b - e ) + 1 03(9c - d) + 1 02(9d - c ) + 1 0(ge - b) + 80 = 0 , whence 8999b + 890c = 991e + lOd - 8 . B ut maximum ( 9 9 1 e + 1 0d - 8) = 900 1 , so b = O. Then, 890c = 991e + 1 0d - 8 , e ntailing e = 8 . Thence , 89c - d = 792, so c = 9, d = 9. N = 109989. S ay, a = 2 , k = 4, f = 8 . Then 1333b + 130c + 1 = 20d + 332e . But maximum (20d + 332e ) = 3168, so b < 3. Also, ( 1333b + 1) must be even, so b must be odd, hence b = 1 . T h e n , 1 0d + 166e = 6 5c + 6 6 7 . But maximum (65c + 667) = 1352, so e < 8 . Also, we require ( 1 0d + 166e ) > 666. So e > 2. But ( 1 6 6e - 667) must be divisible by 5 , hence e = 7. Thence , 13c - 2d = 99, so d = 9, c = 9. N = 2 1 9978. So 2 1 9978 · 4 = 8799 1 2 .

96. A MATTE R O F TIME

Say he went out at z hours , w minute s . The h o u r h and was at (60z + w )/12 minute s , minute h and at w . Say h e returned a t x hours, y minute s . T h e hour h and w a s a t (60x + y )/12 minute s , minute hand at y . Then, y = (60z + w )/ 1 2 , and (60x + y )/12 = w + 2 . Co mbining th e two e q u ation s : 60z = 143w + 2 8 8 - 720x . B ut y i s an integer, so w = 12k, say, k being some integer. Then 5z + 60x = 143k + 24. B ut k i s an integer, and obviously k < 5 , so k = 2 , m aking z + 12x = 6 2 . T h e n , since z ::::; 1 2 , w e h ave z = 2 , with x = 5 . Thenc e , y = 12, a n d w = 2 4 . H e went o u t a t 2 : 24 p . m . , returning a t 5 : 1 2 p.m.

SOLUTIONS TO T EASERS

83

97. THE POSTE R

z - - - I

y I I I

I '--__.& z

_

_

_

(x

- 2z)

_

_ _

_

_

_

�

_

_

z

_

J

(x - Z)2 = y2 + Z2, whence z = (x2 - y2)/2x . Then, x 2z = y2/X , so w 2 = (x - 2Z)2 + y2 = (y4 + x2y2)/X2, y v'(x2 + y2) m aking w = . -

x

Here we h ave y = 120, w = 136, so 136x = 120 v'X2 + 14400, whence x The poster was 225 cm long.

=

225 .

98. AT THE CASINO

This problem i s a vari ation on the classic problem of the odds against two people in a certain group h aving the same birthd a y : " Say there are n people in a roo m . What is the prob­ ability th at at le ast two of them h ave the same birth­ day in the year (i.e., same d ate in the year, not nec­ essarily the same ye ar)?" Ignoring Fe bru ary 29th, a possible leap year birt h d ay, the exact probability is given by: 1 -

�

365 n · 1 (365 - n)

In our roulette vari ation on the theme we h ave 37 num-

84

CHALLENGING MATHEMATICAL TEASERS

bers corresponding to 365 d ays, and n spins correspond­ ing to n people . Odds of " 3 to 1 against" means proba­ bility 1/4 . 1 = 4" approXIm ate I y. S 0, we h ave 1 � n . 37 , (37 n) _

.

With n = 4 we h ave probability 0.1 5424 approximately, with n = 5 , prob ability i s approximately 0.24566, with n = 6, prob ability i s approximately 0.3476 1 . Hence there h ad been 5 spins o f t h e wheel. 99. NO DIRECT ROAD

Cos () = (x 2 + 64 - 2 5)/16x = (x 2 + 39)/16x . Sin2 () = 1 - x 2 + 39 2/25 6x 2 = - (x 4 - 1 78x 2 + 1 52 1 )/256x2. 178x2 - X 4 - 1 5 2 1 S i n () = 1 6x Also, cos (60° - () = (x 2 + 64 - 9)/16x = (x2 + 55)/16x . But, cos (60° - () = cos 60°cos () + sin 600sin (), so (x 2 + 5 5)/16x = (x 2 + 39)/32x + V3( 1 78x2 - X 4 - 1 5 2 1 )/32x , whence X4 - 98x 2 + 2401 = 0, and x = 7. So the vil lages were 7 miles apart. 1 00.

A NE ST EGG

Say the mother's age was m years when M ary was 1 year old. Then, when M ary's age was n years, Geoffrey deposited n (m + n - 1) dollars.

SOLUTIONS TO TEASERS

85

Tabulate for the first few years : Accu mulated Deposits Deposit Year 1 m m + 0 3m + 2 2m + 2 2 3 6m + 8 3m + 6 4 10m + 20 4m + 12 It can be seen that when Mary's age was n years, the total deposits would amount to n(n + 1)(3m + 2n - 2)/6. So, n(n + 1)(3m + 2n - 2) = 888 · 6 = 24 . 32 . 37. n and (n + 1) are consecutive integers , so e ach must be a factor of 24 . 32 . 37. Also we m ay assume m > 1 2 , so n < 10. Then there are only 3 possible values for n giving: n(n + 1) = 2 · 3 3·4 8·9 m aking m = 8 86/3 440/3 2 0 Hence h e r mother was 1 9 years old when M ary was born .

AL PHAMETICS 1.

L I N G K W O N G L E E L I K E E L I C H E E

2.

Little crosses i ndicate digits you will h ave to identify. There is no " re mainder."

x x x ) x x x x x x ( x x x x x

8

x x x x x 8 x x x x x x x x x

3.

N I N E L E S S T W O

Two is the only even pri me number, and here our TWO is truly pri m e !

S E V E N

4.

A N D S O T O B E D S A Y D P E P Y S

Long, long ago i n " y e o l d e E n gland " ! B u t don't forget our PE PYS was odd.

x

88 5.

CHALLENGING MATHEMATICAL TEASERS Well, wel l ! The STEAKS must be odd.

K A T E'S K I T T E N H A T E S T H E S E S T E A K S

6.

K N a C K

7.

K N a C K I T'S T H E P O S T I

C O P P E R K E T T L E S

E

8 . Little crosses indicate digits you will h ave to identify. There is no "remainder."

P E T E R P O P P L E P E D D L E S

x x x ) x x x x x x x x ( x x 7 x x x x x x x x x x x x x x x x x x

9.

They're beautiful beasts with their long t apere d horn s, and i n fact this i s a prime ORYX !

O R Y X O R Y X R U N O N K A R O O

10.

That's where he lived, of cours e . But Holmes never solved probl e m s like this !

R E S T B R E A K A T B A K E R S T R E E T

11.

P A S H A'S H U B B L E

12.

P R E T T Y

B U B B L E

P O L L Y P A R R O T

B U B B L E S

R E P E A T S

x

ALPHAMETICS 13.

H ardly a n ational le ague pitcher!

fl H Y O H W H Y L O B L A D Y

14.

If they rattle it's ripe ! So our PI PPI N will not be odd.

S H A K E S H A K E P I P S I N P I P P I N

15.

1 6.

We have only the one 7 i n this, and it's the 7 you see. No " remainder."

L I L Y L I K E S S I L K Y S I L K Y

x x x ) x x x x x x ( 7 x x x x x x x x x x x x x

Sleek and sheer, of course, but the N Y L O N S will not be odd !

N Y L O N S

1 7.

P E T E R

18.

S U S I E

P A P E'S

S E E S

A P A P E R L I T T E R

A B L U E

P I C K E R

B L U E B E E T L E

89

90 1 9.

CHALLENGING MATHEMATICAL TEASERS Here we squ are a 4-digit n u m ber. The little crosses indicate digits to be identifi e d .

x x x x x x x x x x x x x x x x x x 3 x x x x x x x x x x x x x 3 x

20.

D I V E R D I V E S D E E P I N

It could be dangerous, what with weeds and other s n ags . B ut that D r V E R is truly prime !

R I V E R

21.

They're morons, of course ! But it should be a pri me STREAK.

S E E K O O K Y K O O K S S T R E A K

22.

24.

P E T E R

23.

A D A M

P I P E R L I K E D R E D

S A Y S

R E D

A D A M

P E P P E R

M A D A M

x x x x x x 7 x x x x x x x x x x x x x x x

Y E S I'M

We h ave only the one 7 in this, and it's the 7 you see .

x x x x x x x x

25.

A tougher variation on the old classic SEND MORE MONEY theme !

M O N E Y P O P S O M E M O N E Y P L E A S E

ALPHAMETICS 26.

A L L A R U D D Y R U D D Y

There 's no real muddle here, but you should find this M U D D L E odd !

M U D D L E

27.

29.

W H E N

28.

B O N G O

I N R O M E

B O N G O B O N G O

B E A

O N T H E

R O M A N

C O N G O

J E L L Y J E L L Y A L L W I L D

That J E L L Y i s quite special ! In fact, it is truly pri m e .

W I B B L E W O B B L E

30.

A "toughi e . " No "remainder."

T H E ) F I S T S ( M A Y F L Y

x x x x x x x x x x x x x x -

31.

H O L M E S S O L V E D M O O D

-

-

-

There were many of that type , but the M U R D E R S must be truly odd !

M U R D E R S

32

.

M E I N

33.

E S K I M O

P A P A L I L L I

M I S S M I S S E S

P A L M E R

I C E

T R I L L E D

I C I C L E S

91

92 34.

CHALLENGING MATHEMATICAL TEASERS F E M A L E S

35.

C R A B B Y

A T L A S T

C R A B B Y T A B B Y

D E F E A T

M A Y

E V E R E S T

S C R A T C H

36.

M E N A N D

x x x x x x x x x W O M E N

37.

S U S I E'S S I S T E R S I S S I

E

S K I S I N S U S I E'S

With all th at snow B I KI N I S must be odd ! T h at would m ake it easier to solve, but you really do not need the inform ation.

B I K I N I S

38.

40.

C H E F

39.

O L D

F R I E D O R D E R F O R

S A L T T O L D T A L L

F A T H E R

T A L E S B E A D Y E Y E D B L U E B U D G I E G U G G L E S

SO L UTI ONS TO AL PHAMETICS 1.

L I N G K W O N G L E E L I K E E L I C H E E

In this and in many alphametics it is convenie nt to use the symbol - me an­ ing " e nd s with." For example, 8 5 - 5. Al so, if the letter 0 appears in the l ay­ out, the n u meral 0 should be spelled out as " z ero . "

Here we h ave L = 1 , and I = zero. Then in the LWI column, L + I = 1 , so there i s no carry from that column. Hence K = 9 . W e h a v e 2G + E - z e r o , he nce E m u s t be eve n . But, from the N N E E column, with E even we need an even c arry from GGE E , so 2G + E = 20. Now tabulate for possible E value s , with corre sponding values for G and other letters. Duplications or otherwise un acceptable values should be struck out or omitted. 8 4 E 6 6 7 8 G 2N + E - 8, so giving carry

N

I + L + K + carry

gi vi ng carry

0 with H

L + I + carry

2 1

7 2

5 2

11

12

12

5 6 1

6 7 1

3 5 1

2 4 1

2

2

2

2

94

CHALLENGING MATHEMATICAL TEASERS

W = - 3 with C = 5 H e nce we h ave uniquely: 1028 + 93628 + 144 + 1 0944 1 05744. 2. x x x ) x x x x x x ( x x x x x

8

x x x x x 8 x x x x x x x x x

x

=

Re the 2nd product, the divisor < 125. Obviously, the quotient i s 989, so the divisor is greater th an 999/9, i.e., > 1 1 I .

Con s idering the result of su btracting the 2nd product, we see it must start with 88 or 98. I f the 2nd product starts with 88, as a multiple of 8 it must be 888, impossible because the divisor > 1 1 I . S o the 2nd product starts with 98, and must b e 984. H e nce the divisor i s 123, and we h ave the complete cal­ culation as : 1 2 1 647 -;- 123 = 989. 3. N I N E LESS TWO SEVEN = 6862 42 1 1 953 12026, with 9 and 8 interchange able. 4. AND SO TO B E D SA YD PEPYS 9287 1 0 1 89.

=

237 94 64 507

5. KAT E ' S KITTEN HATE S THESE STEAKS = 2 5 1 93 2 0 1 198 75 1 93 1 7939 3 1 9523.

6. KNOCK KNOCK IT'S T H E PO STl E 594 903 1 74953.

=

86728 86728

7. PE T E R POPPLE PE D DL E S COPPER KETT L E S 10705 1 6 1 1 9 0 1 044903 8 6 1 1 05 2077903. 8. 1 00079 1 6 -;- 1 2 4

=

=

80709.

9. ORYX O R YX RUN O N KAROO 1 0944.

=

4987 4987 925 45

1 0. R E ST B R E AK AT BAKE R STR E E T 43 64275 1 3 5773.

=

5 7 1 3 65742

SOLUTIONS TO ALPHAMETICS 1 1 . PASHA'S H U B B L E B U B B L E B U B B L E S 97 1 1 20 1 7 1 1 20 1 7 1 1208. 1 2 . PR E TTY PO L L Y PAR R O T R E PEATS 79003 781 196 1 5 75862. 13. 314

x

21

=

+

568968

7 1 5663

=

6594.

1 4 . SHAKE SHAKE PIPS I N P I PP I N 1 3 1 6 30 1 3 1 130. 1 5 . 1 00536

=

142

=

=

64892 64892

708.

1 6. L I LY L I KE S S I L KY S I L KY N Y L O N S 53482 23540 23540 1 0 5 9 1 2 .

5350

=

1 7. PE T E R PAPE ' S A PAPE R L I TTE R P I C KE R 36 1 64 35367 5 35364 2 9 1 164 398064. 1 8 . S U S I E S E E S A B L U E B L UE B E E T L E 8008 2 1 390 1390 100630. 1 9. 3 1 94

x

95

3 1 94

=

=

89840

=

1020 1 636.

20. D I V E R D IVE S D E E P I N R IVER = 45 1 79 45170 4772 5 8 95 1 79, with 0, 2, 8 interch ange able . 2 1 . SE E KO O KY KOO KS STREAK 1 53487.

=

1 44 76672 7667 1

22. PE T E R P I PE R L I KE D RED R E D PE PPE R 1 2628 1 3 1 2 8 93724 824 824 1 2 1 128. 23. ADAM SAYS Y E S I ' M ADAM MADAM 324 9 1 5 0 5 1 1 5 05 1 . 24. 1 238

x

8079

=

=

=

5 0 5 1 4534

=

963 1 4 262

10001 802.

25. MONEY POP S O M E MONEY PL E A S E 8691 963 1 4 20 1 5 8 1 . 2 6 . A L L A RUDDY R U D D Y M U D D L E 62550 1 2 5537.

=

27. WH E N I N ROME BE A R O MAN 3 1 0637.

9457 87 1 06 5 25

=

433 4 62550

28. B ONGO B O NGO B ONGO ON THE CONGO 297 1 9 29719 97 465 897 1 9 .

=

29719

29. J E LL Y J E L L Y ALL W I L D W I B B L E WOB B L E 39227 39227 522 1024 1 06629 1 86629.

=

96

CHALLENGING MATHEMATICAL TEASERS

30. 97848

+

453

=

216.

3 1 . H O L M E S SO LVE D M O O D M U R D E R S 934820 1 330 1 5 70279.

634 129

=

32. M E I N PAPA L I L L I PAL M E R TR I L L E D 9494 73773 947680 1 037782. 33. E S K I M O M I S S M I S S E S I C E I C I C L E S 6 1 88 6 1 8878 1 4 7 1 4 1 4378.

=

=

6835

789 165

34. F E MA L E S AT LAST D E FEAT EVE R E ST 349 7 1 4 8 76 1 786 543476 4042486.

=

35. CRAB B Y CRA B B Y TAB B Y MAY SCRATCH 964552 964552 34552 742 1 964398. 36.

125

x

753

=

=

94 1 2 5 .

3 7 . S U S I E ' S S I ST E R S I S S I E SKI S I N S U S I E ' S B I K I N I S = 9 1 9879 989470 989987 9289 8 5 9 1 9879 3828589. 38. CHEF FR I E D O R D E R FOR FAT H E R 96356 1 96 120856. 39. O L D SALT T O L D TALL TAL E S 1099 1 0978.

=

=

785 1 16453

394 8091 1394

40. B E ADY E Y E D BLUE B U D G I E GUGG L E S 3635 9803 905 1 43 1 0 1 1 837.

=

93256

APPEN D ICES APPENDIX I: DIOPHANTINE E QUATIONS

Diophantos, who lived during the third century A.D., was probably the first mathematician to make an exten­ sive stu dy of the types of equ ations that we now associate with his name. The s alient fe ature of what are known as " diophantine equ ations" is that they involve two or more unknowns that in almost all cases repre sent integers . The simplest such e q u ation is known as the " Pell E q u a­ tion" : X 2 - NY 2 = 1 where N is a non-squ are integer. There will be an infin ite number of (X, Y) pairs of i ntegral values sati sfying this equation. We have initially X = 1, Y = O. H aving determined the next gre ater pair of (X, Y) value s , all further integral solutions are given by: X n+ 2 = 2aX n+ 1 - X n Y n+ = 2aY n+ 1 - Y n 2 where a is the value of X in that " next gre ater" solution. For example, the first two pairs of values for the equa­ tion X 2 - 3Y2 1 are (X, Y) = ( 1 , 0) and (2 , 1), m aking a = 2. Then we can write down all successively gre ater pairs of values at sight : X = 1 2 7 26 97 etc. Y = 0 1 4 15 56 etc. Then there i s the more general form X2 - NY 2 = e , where e is some positive or negative integer other t h an + 1. This may include, as a particular case, X 2 - NY 2 = - 1 . For any p articular value of N there will be integral solutions only for certain values of e , and vice versa. The =

98

CHALLENGING MATHEMATICAL TEASERS

theoretical problem of d etermining such compatible val­ ues for N and e is complex. B ut in practice the nece ssary confirm ation as to whether or not integral solutions exist for any given N and e can be found by testing over a limited range of valu e s for Y. Let us assume that we h ave ascertained that there is an integral solution for X2 - NY 2 = e , for particular val­ ues of N and e. And let u s assume that this be the integral solution that involves minimal value s X = x , Y = y. Also, let X = m, Y = n, be any i ntegral solution of the Pell equ ation X2 - NY 2 = + 1, the N having the same value as in the origi n al e q u ation. Then, X 2 - NY2 = e = x 2 - Ny2, and m2 - Nn2 = 1, so X2 - NP = (x2 - Ny2)(m2 - Nn2)

x 2m 2 + N2y2n2 - Nx2n2 - Ny2m2 x2m2 ± 2Nxymn + N2y2n2 - Nx2n2 =+= 2Nxymn - Ny2m2 = (xm ± Nyn)2 - N(xn ± ym)2 E quating coefficients in the origi nal X2 - NY 2 = e, we h ave X xm ± Nyn , Y = x n ± ym, the sign s " + " or " - " being the s ame i n e ach. Thence, substituting for m and n any p air of values that satisfied the equ ation m2 - Nn2 = 1 , we obtain solutions for X 2 - NY2 = e . A n example will m ake the procedure cle ar. Say X 2 3 Y 2 = - 1 1 , the minimal i ntegral solution of which is X = 1 , Y = 2. Then other solutions are given by: X = m ± 6n, Y = n ± 2m But ( _X)2 = ( +X)2, and ( _ y)2 = ( + y)2 , so we may write =

=

=

-

this ge neral solution more ne atly as : X = ± (m ± 6n), Y = ± ( 2m ± n ) where the internal sign s agree , a n d m a n d n satisfy m 2 3n 2 . = 1 . W e c a n t abul ate succe ssive values o f m a n d n , as al­ re ady explained, writing down corre sponding values of X and Y: m = 1 2 2 7 7 26 26 etc. n = 0 1 1 4 4 15 1 5 etc. X = 1 4 8 1 7 3 1 64 116 etc. Y = 2 3 5 1 0 1 8 37 67 etc. This has been a very brief outline of the theoretical appro ach to the si mplest types of diophantine equations. It may help the re ader to understand what is involved in some of the detailed solutions to probl e m s in this collec­ tion.

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APPENDIX I I : THAT RE MAINDER B USINESS

The m athe matici an calls it Congruence The ory, which may sugge st something rather complicate d . B ut here we look into some eleme ntary aspects of this important branch of Number Theory i n terms of the familiar re­ mainder that may re sult fro m the process of si mple divi­ sion. Congrue nce The ory, propounded and developed by the gre at Carl Gauss when he was still i n his teens, is entailed in many popul ar-type mathe matical teasers. To under­ stand and apply its pri nciples in the solving of such prob­ lems does not i nvolve delving deeply into abstru se theo­ retical con siderations. I n this brief survey the emphasis will be on practical applications rather than on theory. We shall be dealing only with integers, i . e . , whole n u m­ bers. It is esse ntial, of course , to use and understand the special notation. Dividing 23 by 7, say, we h ave re mainder 2. This would be shown a s : 23 == 2(mod 7 ) which is re ad in word s as "23 i s congruent t o 2, t o modulus 7." In other words, "23 le aves a re mainder of 2 when divided by 7." Stretching our concept of re mainder, ob­ viously an infi nite number of integers must be congruent to 2, to modulus 7 . For exampl e : 30 == 2(mod 7) 2 == 2(mod 7 ) - 5 == 2(mod 7) In fact, all such integers are solutions of the e q u ation x = 7k + 2, where k i s any i nteger (positive, negative or zero). A few m ore examples will famili ari z e the re ader with these ideas. Now we have seen that, to modulus 7, the integers 23 and 2 are equivalent. I f we m ultiply one side of our orig­ inal congrue nce equation by 2 and the other side by 23, the new congruence must then be v alid : 46 == 46(mod 7 ) 5 2 9 == 4(mod 7) The first of these needs n o comment, but notice the sec­ ond for future refere nce . Clearly, 1 2 == - 2(mod 7), and for exampl e : 1 9 == - 2(mod 7) - 2 == - 2(mod 7 ) Since 1 9 and - 2 are equivalent, to modulus 7, we may m ultiply one side of the first by 19, and its other side by - 2, to obtain a new congrue nce e q u ation: 3 6 1 == 4(mod 7) Note here that 3 6 1 1 92, and e arlier we h a d 529 = 232• =

100

CHALLENGING MATHEMATICAL TEASERS

N ow, say we have the equation x 2 - 4 = 7y , where x and y are integers. This may be writte n as x2 == 4(mod 7), which is satisfied by the ge neral integral solution x 7k ± 2, i . e . , x == ± 2(mod 7). If only positive values are required we have an infinite seque nce of such values : 2, 5 , 9, 12, 1 6 , 19, 23, etc. I n solving some problem we might derive the rather more complex equation x 2 - 6x + 6 = 1 1y , say. This be­ comes (x - 3)2 1 1y + 3, so (x - 3)2 == 3(mod 1 1). B ut, 25 == 3(mod 1 1 ), hence we have (x - 3)2 == 25(mod 1 1), so (x 3) == ± 5(mod 1 1). Thence x == 8 or 9(mod 1 1), making the general solution x = 1 1k - 3, or x = 1 1k - 2. Hence p articular solutions are 8 , 9, 19, 20, etc . , with correspond­ ing values for y. At this poi nt it is sugge s te d that the reader write down a few typical equations of similar type, and derive the general integral solutions. It will probably be found th at some have no integral solutions. Full discussion of such cases must be outside the scope of this survey. However, s ay we have an equation X 2 == a(mod m), equivalent to X mk + a . Then, if there is an integral value of k that will make the right-hand side a perfect square, the smallest such value must lie within the range of integral values k = 0 to k m/2. So , to find the required perfect square (if it does exist), it is only necessary to test for values of k within that range. So far we h ave considered only examples with prime moduli. Now we t ake a case where the modulus is a power of 2. Say X 2 == l(mod 8). Then we need test only for k < 5 in X 2 8k + 1 . We find X == ± l(mod 8) or X == ± 3(mod 8). The first few positive valu es of X are therefore 1 , 3, 5, etc., and we see t hat those two different solutions do in fact comprise all the odd numbers . Hence they can be combined as a si ngle solution, X == ± l (mod 4). I ndeed the ge neral solution of X 2 == l(mod 2n) for moduli that are powers of 2 is X :5 ± l(mod 2n - 1): an important point to re member. Cases of more complex composite moduli must also be considere d . In such cases the modulus must be broken d own into its distinct pri me factors (or powers thereof). Say, for example, that X 2 == l(mod 135). This implies X 2 == l(mod 5) and X 2 == l(mod 27), so we must fi n d solutions in X t h at will satisfy both congruence e q u ations. ==

==

=

==

==

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In ge neral, to d e al with a composite modulus we first set up the requi site number of subsidi ary congruence equation s, depending on the number of pri me factor ele­ me nts i n the modulus. E ach of those su bsidi ary e q u ations must be solve d, using the methods that have been out­ lined, and finally those separate solutions will h ave to be coordinated to give the ge neral solution that will cover all of the m. For example, say X2 == l 1(mod 35), the congruence equa­ tion that re sults from X 2 - 1 1 = 35Y. Here the modulus is 5 ' 7, so X2 == 11 == l(mod 5), and X2 == 11 == 4(mod 7), the distinct solutions being X == ± l(mod 5) and X == ± 2(mod 7). The simplest method for co-ord i n ating those two dis­ tinct solutions follows . It will be given in detail because its applic ation is i nvolved so ofte n i n solving problems that e ntail congruence equations. Say X = 5u + a = 7v + b , where u and v are integers , with a = ± 1 and b = ± 2 (the + and - signs being inde­ pendent). Dividing through by 5 , it is seen t h at (2v - a + b)/5 must be an i nteger. Thence, (v - 3a + 3b )/5 m u st be an integer, say t . So v = 5t + 3a - 3b . B ut, X = 7v + b , so X = 35t + 2 1a - 20b , which implies X == [21a - 20b ](mod 35). N ow we assign their values to a and b, getting X == [ ± 2 1 ±40](mod 35), the + and - sign s being mutu ally inde­ pe ndent. The nce X == ±9 or ± 1 6(mod 35). So the required ge neral solution of X 2 - 11 = 35Y is X 35k ± 9 or X = 35k ± 1 6 . The first few positive valu e s for X being: 9, 1 6 , 19, 26, 4 4 , e t c . If the basic equ ation was derived i n the solving of a popul ar-type "te aser," there would be some speci al stipulation that would pinpoint one or more particular values. Where the modulus has t hree or more different prime factor elements, precisely the same method is used. H av­ ing coordinated two of the subsidi ary congruence solu­ tion s, the re sulting solution is coordinated with the third, and so on. Try solving X 2 - 4 = 1 0 5Y, as a si mple example. =