Journal of Recreational Mathematics, Vol. 35, No. 1, 2006

J. RECREATIONAL MATHEMATICS, Vol. 35(1) 1-4, 2006

SOME RESULTS OF AN INVESTIGATION OF UNSOLVED PROBLEM #1079

CHARLES ASHBACHER Charles Ashbacher Technologies 5530 Kacena Ave. Marion, IA 52302 e-mail: [email protected]

In Volume 14(2) of Journal of Recreational Mathematics [1], Harry Nelson posed Problem #1079: What factorial has the largest percentage of zeros in its decimal representation?

A partial solution was published in a later issue [2], where the number of zeros in the factorials was computed up through 743!. The fact that the highest percentage of zeros was recorded for 7! (50%) and 12! (44.4%) was used to argue that they are the solutions. There appears in the statement: The number of terminal zeros approaches 10% and, as expected, the number of internal zeros approaches 10%. On this basis, it seems reasonable to assume that the number of zeros in the large factorials averages about 20%. A reasonable answer to the stated problem is: 7! has the largest percentage (50%) of zeros in the decimal expansion.

The desire to resolve the original question and to investigate the claims of the statement led me to engage in some further explorations. Using the BigInteger data type in the Java programming language, a program was written to determine the percentage of zeros in factorials. A plot of the percentages of zeros in the factorials from 1 through 6000 appears in Figure 1.

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2 / ASHBACHER

Figure 1.

Note that there is a slow but clear trend of a decreasing percentage of zeros in the factorial. To give greater clarity to this trend, the mean and standard deviation of the percentage of zeros for ranges per thousand are given in Table 1. Note once again the clear trend in the decreases in both the mean and standard deviation in the percentage of zeros. Note: The range 1–1000 was not part of this analysis, as can be seen from the chart, the stable behavior of decline does not appear in the smaller values of N. Therefore, it is a safe conclusion that the percentage of zeros in factorials does not approach 20% as stated in the statement in the “solution” to the original problem, but is in fact much less. In looking at the trend in the graph of Figure 1 and the data in Table 1, a more reasonable approach is 16%. To determine which location of zeros, “internal” (defined as a zero to the left of the rightmost nonzero digit in the factorial) or “terminal” (defined as to the right of the rightmost nonzero digit) is responsible for this, the program was modified and rerun. The graph of the percentage of terminal zeros appears in Figure 2. The mean and standard deviation of the percentage of trailing zeros based on the same ranges of a thousand at a time is given in Table 2. The graph of the percentage of internal zeros is given in Figure 3. The mean and standard deviation of the percentage of internal zeros based on the same ranges of a thousand at a time is given in Table 3. Assuming that the digits in a factorial are from one position to the right of the leading digit and one position to the left of the rightmost nonzero are randomly distributed, then this trend in the means is reasonable. As the impact of the two digits that cannot be zero is lessened, the percentage of internal zeros will slowly rise. Also note that the standard deviations of both classes of zeros also decline

UNSOLVED PROBLEM #1079 / 3

Table 1. Range

Mean

Standard Deviation

1000-2000

0.181888

0.005106

2000-3000

0.175621

0.003671

3000-4000

0.172037

0.002879

4000-5000

0.169768

0.002522

5000-6000

0.167863

0.002255

Figure 2. Table 2. Range

Mean

Standard Deviation

1000-1999

0.090864

0.002772

2000-2999

0.084045

0.001422

3000-3999

0.080202

0.000904

4000-4999

0.07749

0.000676

5000-5999

0.07546

0.000526

over time. This is also expected as the slight increases and decreases in zeros when specific number are multiplied into the factorial will tend to be less significant as the numbers get larger. From the data here, it is clear that 7! = 5040 has the largest percentage of zeros and that part of the problem should be considered closed.

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Figure 3. Table 3. Range

Mean

Standard Deviation

1000-1999

0.091023

0.004653

2000-2999

0.091576

0.003427

3000-3999

0.091836

0.002759

4000-4999

0.092278

0.002437

5000-5999

0.092402

0.002195

The final question to consider deals with the percentage of trailing zeros in factorials as N gets larger. Therefore, I will close with the following question: The means of the percentages of trailing zeros in Table 2 demonstrates a steady decline as N gets larger. What is the limit of this percentage as N goes to infinity?

References 1. H. L. Nelson, Problem 1079, Journal of Recreational Mathematics, 14:2, p. 142, 1981-82. 2. H. L. Nelson, Partial Solution to Problem 1079, Journal of Recreational Mathematics, 15:2, pp. 156-157, 1982-83.

J. RECREATIONAL MATHEMATICS, Vol. 35(1) 11-14, 2006

MAGIC TESSERACT CLASSES HARVEY D. HEINZ 15450 92A Ave. Surrey, BC, V3R 9B1, Canada e-mail: [email protected] MITSUTOSHI NAKAMURA 3-24-8 Midorigaoka Morioka, Iwate, Japan e-mail: [email protected]

Introduction The late John Hendricks was a prolific investigator of magic squares, cubes, and higher dimensions [1]. As early as 1968 [2-5], he was applying descriptive names to certain types of magic hypercubes. By the late 1990s, he had developed a unified classification system for magic hypercubes that was consistent for all dimensions! This was first introduced in a formal manner with an article in JRM in 2004 [6]. A modified version of this classification system, as applied to magic cubes, was required due to discovery by one of us, Mitsutoshi Nakamura, or a previously unknown type of magic cube [7]. After this discovery, Nakamura turned his attention to magic tesseracts and has since defined all 18 classes of these 4-dimensional hypercubes and constructed examples of most of them [8, 9].

More on Class Names Here we will present the universal classification system from a different viewpoint than previously discussed. John Hendricks originally coined the term n-agonals for the lines of a magic hypercube that was of dimension n. The term

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12 / HEINZ AND NAKAMURA

pan-n-agonals referred to all the main diagonals and all broken lines of length n that were parallel to these main diagonals (pan means all). In the case of a square (dimension 2), these lines are called the main diagonals and pan-diagonals. For a cube they are called tri-agonals and pan-triagonals (and sometimes space diagonals). And so on for the higher dimensions. Notice that when you move along a diagonal, 2 co-ordinates change, along a triagonal, 3 co-ordinates change, and in general, along an x-agonal, x co-ordinates change. With this in mind, it is consistent to consider the orthogonal rows (those parallel with the edges of the magic hypercube), to be 1-agonals because only one co-ordinate changes as we move along the row. If we wish to collectively refer to all the agonals in a magic hypercube, the term r-agonals (or pan-r-agonals if appropriate) is often used. In order for a hypercube to be magic, all 1-agonals, and the main (corner-to-corner) n-agonals, must sum to a constant, which is usually referred to as S. S = m(1 + mn)/2, where m is the order of the hypercube (length of each side) and n is the dimension (square, cube, tesseract, etc.). The features just described are those of the lowest class in each dimension of hypercube. It is called simple magic, a term that has been around for many years. These features are usually unmentioned but are a required part of each of the other (higher) classes. In describing his classification system, John Hendricks never specifically mentioned that the classes were determined by the various combinations of r-agonals or pan-r-agonals. However, by the time of the discovery of the sixth class of magic cube, it was very obvious that this, in fact, was the basis of this whole method of classification. This idea has been used, by tradition, in the naming of the two classes of magic squares (simple and pandiagonal), and recently, six classes of magic cubes. Nakamura, during his discovery and naming of the 18 classes of magic tesseract, has carried on this tradition. The Magic Tesseract Classes Table 1 ranks the 16 classes from lowest (fewest summations) to highest. Shown are the summations required for each of the four types of r-agonals as well as the total. As usual, m indicates the order of the tesseract. More information on how these values are derived and are available from Heinz’s Tesseract Math page [10]. Note that most tesseracts will contain more summations than required for the particular class, but not sufficient to qualify for the next higher class. This table shows the required summations. There are often other patterns that also sum to S. A reminder that the nasik (top class, or perfect) tesseract contains many lesser nasik magic objects. All 6m2 squares and 4m cubes in a nasik tesseract are nasik magic! This features applies to all dimensions of nasik magic hypercubes!

MAGIC TESSERACT CLASSES / 13

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And a final humbling note. John Hendricks is credited with producing the first set of all 58 order-3 magic tesseracts and the first nasik magic tesseract (as well as a great deal of work dealing with the relationships between the various dimensions of magic hypercubes). In the last 10 or so years, many other investigators have contributed to this field. However, in 1905, C. Planck produced several order-3, and order-8, and one of the 256 planes required for an order-16 nasik tesseract (he called these octahedroids) [11, 12]. Summary We have shown how the magic tesseract may be organized into 18 classes as determined by the conditions of the 2, 3, and 4-agonals. This is an extension of John Hendricks Universal Classification System as originally demonstrated for magic squares and cubes. Dimension-2 hypercubes have two classes, dimension-3s have six classes, and dimension-4s have eighteen classes. Therefore, we can expect higher dimension magic hypercubes to contain many more classes. The number of classes for a particular dimension n is 2 × 3n-2. References 1. In Memoriam: John Robert Hendricks, Journal of Recreational Mathematics, 34:1, pp. 80-81, 2005-2006. 2. J. R. Hendricks, The Pan-4-agonal Magic Tesseract, American Math Monthly, 75:4, p. 384, 1968. 3. J. R. Hendricks, The Pan-3-agonal Magic Cube, Journal of Recreational Mathematics, 5:1, pp. 51-52, 1972. 4. J. R. Hendricks, Magic Tesseracts and n-Dimensional Magic Hypercubes, Journal of Recreational Mathematics, 6:3, pp. 193-201, 1973. 5. J. R. Hendricks, Pan-n-agonal in Hypercubes, Journal of Recreational Mathematics, 7:2, pp. 95-96, 1974. 6. H. D. Heinz and J. R. Hendricks, A Unified Classification System for Magic Hypercubes, Journal of Recreational Mathematics, 32:1, pp. 30-36, 2003-2004. 7. H. D. Heinz, Hypercube Classes—An Update, this issue pages 5-10. 8. M. Nakamura, http://homepage2.nifty.com/googol/magcube/en/classes.htm. 9. H. Heinz, http://members.shaw.ca/tesseracts/t_classes.htm. 10. H. Heinz, http://members.shaw.ca/tesseracts/t_math.htm. 11. C. Planck, The Theory of Path Nasiks, printed for private circulation by A. J. Lawrence, Printer, Rugby (England), 1905 (available from The University Library, Cambridge). 12. W. S. Andrews, Magic Squares and Cubes, by Dr. C. Planck, Open Court Publ., pp. 363-375, 1917. Re-published by Dover Publ., 1960 (no ISBN; Dover Publ., 2000, ISBN 0486206580; Cosimo Classics, Inc., 2004, ISBN 1596050373.

J. RECREATIONAL MATHEMATICS, Vol. 35(1) 15-19, 2006

TWO OBSERVATIONS ON UNSOLVED PROBLEM #1046 ON PRIME CIRCLES OF {1, 2, . . . , 2m}

MICHAEL A. JONES Mathematical Reviews 416 South Fourth Street Ann Arbor, MI 48103 e-mail: [email protected] LESLIE CHETEYAN Dept. of Mathematical Sciences Montclair State University Montclair, NJ 07043 e-mail: [email protected]

ABSTRACT

An unsolved problem from the problem section of this journal is whether or not there exists a prime circle for the set {1, 2, . . . , 2m} for every m [1, 2]. We show that if a restricted version of this problem in which the odd numbers are arranged in numerical order always has a solution, then the Twin Prime Conjecture is true. Further, we re-state the original problem in terms of finding a Hamiltonian cycle for a graph.

Introduction At the Mathematical Association of America’s MathFest in 2008, Charles Ashbacher, one of the editors of this journal, discussed a number of problems that were introduced in the problem section of this journal, but remain unsolved. 15 © 2009, Baywood Publishing Co., Inc. http://baywood.com

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Ashbacher’s talk included Problem 1046 that appeared in issues from 1982 and 1983 [1, 2]. This unsolved problem was posed by Fitz, and concerns whether, for every m, there exists a prime circle of {1, 2, . . . , 2m}, a circular permutation of the first 2m positive integers such that every adjacent pair sums to prime numbers. This problem also appears in Guy’s book of unsolved problems in number theory [3]. There are two arrangements of the elements of {1, 2, . . . , 8} so that the sums of adjacent pairs sum to prime numbers, as displayed in Figure 1. The sequence that describes the number of prime circles for a given m is A051252 on the Online Encyclopedia of Integer Sequences (O.E.I.S.) [4]. Prime Circles with Odd Numbers in Numerical Order For m = 4, one prime circle has the odd numbers in numerical order (the right circle in Figure 1 with the odd numbers boldfaced). Are there solutions in which the odd numbers are in numerical order for every m? Obviously, answering this question in the affirmative also answers Fitz’s original question. The sequence describing the number of prime circles for this restricted problem is also listed on the O.E.I.S. as A072616 [5]. The O.E.I.S. entry includes the first 30 terms; the first 10 terms are given in Table 1.

Table 1. The First 10 Items of the Sequence a(m) that Describe the Number of Prime Circles of [1, 2, . . . , 2m with the Odd Numbers in Numerical Order m

1

2

3

4

5

6

7

8

9

10

a(m)

1

1

1

1

4

8

2

5

18

2

Figure 1.

UNSOLVED PROBLEM #1046 / 17

In between sessions at MathFest, the authors recognized a relationship between prime circles with the odd numbers in numerical order and twin primes. To motivate the relationship, return to the prime circle with the odd numbers in order given in Figure 1. Beginning with 1 and computing sums of pairs while moving clockwise around the circle, the sums for this prime circle are (in order) 3, 5, 11, 13, 11, 13, 11, and 5. Because the odd numbers are in numerical order, the sums of an even number with the two consecutive odd numbers differ by 2. If these sums are both primes, then they are twin primes. More formally, assume that there is a prime circle of {1, 2, . . . , 2m} for which the odd numbers are in numerical order and the first m even positive integers a1, a2, . . . , am are arranged in the circle such that ak is in between 2k – 1 and 2k + 1 for k from 1 to m – 1 and am is between 2m – 1 and 1. This arrangement appears in Figure 2. For this circle to be a prime circle, then a1 + 1, a1 + 3, a2 + 3, a2 + 5, a3 + 5, a3 + 7, . . . , am-1 + 2m – 3, and am-1 + 2m – 1 would have to be prime, as well as am + 1 and am + 2m – 1. It follows that ak + 2k – 1 and ak + 2k + 1 are twin primes, for k = 1 to m – 1. This observation leads to the following proposition. Proposition: If there exists a prime circle of {1, 2, . . . , 2m} for every m with the odd numbers in numerical order, then the Twin Prime Conjecture is true. Proof: Suppose that, for every m, there exists a prime circle of {1, 2, . . . , 2m} in which the odd integers are in numerical order, but that the Twin Prime Conjecture is false. Then there exists a finite number n of twin primes with lesser prime given by p1, p2, . . . , pn such that pi < pj if and only if i < j. Select m so that 2m – 3 > pn. Because there exists a prime circle with the odd numbers in numerical order, then there exists a permutation a1, a2, . . . , am of {2, 4, . . . , 2m} so that am-1 + 2m – 3, and am-1 + 2m – 1 are twin primes (as in Figure 2). But, am-1 + 2m – 3 is greater than the largest twin prime pn because am-1 + 2m – 3 > 2m – 3 > pn, which contradicts there being a finite number of twin primes. Of course, all it takes is an infinite number of prime circles of {1, 2, . . . , 2m} with the odd numbers in numerical order for the Twin Prime Conjecture to be true. Further, the existence of an infinite number of twin primes does not imply the restricted prime circle result. A Graph Theoretic Restatement of the Prime Circle Problem We consider a graph-theoretic representation of the prime circle problem. Define a graph Gm with vertex set {1, . . . , 2m} and edge set {(i, j) ½ i + j is prime; i ¹ j}. The condition i ¹ j ensures that 1 is not adjacent to itself. For example, the graph Gm for m = 4 is given in Figure 3. It follows that a prime circle of {1, 2, . . . , 2m} is a Hamiltonian cycle of the associated graph Gm. Indeed, the Hamiltonian cycle marked with solid edges in

18 / JONES AND CHETEYAN

Figure 2. A general prime circle of {1, 2, . . . , 2m} with the odd numbers in numerical order.

Figure 3. The graph G4 for the vertex set {1, 2, . . . , 8}. Both solid and dashed lines are edges in the graph. The solid lines are a Hamiltonian cycle or prime circle.

Figure 3 corresponds to the prime circle on the right side of Figure 1. Hence, problem 1046 can be restated as: Is the graph with vertices {1, 2, . . . , 2m} and edges {(i, j) ½ i + j is prime; i ¹ j} Hamiltonian for all m?

Some properties of the graph Gm are apparent by definition. Let p(n) be the number of prime numbers less than n. The degree of vertex i (¹ 1) in graph Gm is

UNSOLVED PROBLEM #1046 / 19

p(2m + i + 1) – p(i + 1) while vertex 1 has degree p(2m + 2) – 1. Further, the graph exhibits a type of symmetry: if vertex i is adjacent to vertex j, then vertex i + 1 is adjacent to vertex j – 1, as long as i + 1 < 2m + 1 and j – 1 > 0. The dashed edges in Figure 3 exhibit the symmetric property for vertices i = 4 and j = 3. References 1. A. Fitz, Problem 1046, Journal of Recreational Mathematics, 14, p. 64, 1982. 2. A. Fitz, Problem 1046, Journal of Recreational Mathematics, 15, p. 71, 1983. 3. R. K. Guy, Unsolved Problems in Number Theory (3rd ed.), Springer-Verlag, New York, p. 160, 2004. 4. N. J. A. Sloane, Sequence A051252 in “The On-Line Encyclopedia of Integer Sequences,” published on-line at http://www.research.att.com/~njas/sequences 5. N. J. A. Sloane, Sequence A072616 in “The On-Line Encyclopedia of Integer Sequences,” published on-line at http://www.research.att.com!~njas/sequences

J. RECREATIONAL MATHEMATICS, Vol. 35(1) 20-22, 2006

BEHFOROOZ-EULER KNIGHT TOUR MAGIC SQUARE WITH U.S. ELECTION YEARS

HOSSEIN BEHFOROOZ Utica College Utica, NY 13502 e-mail: [email protected]

The year 2007 was the 300th anniversary of Leonard Euler’s birth year. He was a great mathematician who has contributed excellent results in mathematics, physics, and other fields. At the same time, this brilliant man spent his valuable time on recreational mathematics, too. He was the first person to introduce the knight tours on the chessboard and the knight tour magic squares. In the literature, we can see several open or closed knight tour magic squares from him. In his birth year celebrations, we witnessed many special lectures about his life and his works. Many books and posters came out. But, I am afraid, I didn’t see anything on his recreational mathematics studies. I picked up one of his closed knight tour magic square from [1, p. 191] and Americanized it and I fixed the following knight tour magic square with 64 U.S. election years from 1788 to 2040. It is worth mentioning here two important points about the knight tour magic squares. From day one, people wanted to know the number of possible knight tours and knight tour magic squares. Very recently, we received the final answers to these questions. By using powerful computers, it has been shown that there are 26,534,728,821,064 distinct closed knight tours and only 140 distinct knight tour magic squares; see [3, 4]. The second mystery was the incompleteness of these magic squares. By using the integers 1, 2, 3, . . . , 64 we have seen many open complete knight tour magic squares with magic sum 260 for all rows, columns, and two diagonals. But there was no complete closed knight tour magic square with magic sum 260. In the closed case, the sum of the rows and columns are 260 but the diagonal sums are two different numbers 256 and 264. For almost 300 years it was 20 © 2009, Baywood Publishing Co., Inc. http://baywood.com

BEHFOROOZ-EULER KNIGHT TOUR / 21

Figure 1. Behforooz-Euler Knight Tour Magic Square with U.S. Election Years.

a dream to have a complete closed knight tour magic square with magic sum 260 for all rows, columns, and two diagonals. The fact is that this will not happen. The reason is so simple and obvious. On the chessboard, the knight piece moves in alternate color cells (from white to black then from black to white or vice versa). So, in any closed knight tour magic square, all the black cells, for example, contain odd numbers and all the white cells contain even numbers. Also, all the cells of one diagonal are white and all the cells of the other one are black. That means, one diagonal contains odd integers and the other one contains even integers. With a simple calculation, we find that 1 + 3 + 5 + . . . + 65 = 1024 and 2 + 4 + 6 + . . . + 64 = 1056. When we divide these two totals by 4, we obtain 256 for the odd diagonal and 264 for the even diagonal, and the conclusion is that we will not see a complete knight tour magic square; see also [2, p. 93]. But the question is why did a great mathematician like Euler not notice this simple argument and why did people for 300 years expect to see a complete knight tour magic square with magic sum 260 for all rows, columns, and two diagonals. Definitely, this problem was not similar to Fermat’s Last Theorem to wait for 300 years and have a proof with more than 100 pages. This is the beauty of mathematics. Sometimes simple justifications are

22 / BEHFOROOZ

not visible and are left to the next generation. Remember that at the end of every research in mathematics we open many doors, roads, and paths for others to start and continue that study. References 1. M. Gardner, Mathematical Magic Show, An MAA Spectrum Book, Washington, DC, 1989. 2. C. Pickover, The Zen of Magic Squares, Princeton University Press, New Jersey, 2002. 3. Available at http://en.wikipedia.org/wiki/Knight’s tour. 4. Available at http://home.freeuk.net/ktn.

J. RECREATIONAL MATHEMATICS, Vol. 35(1) 23-29, 2006

SOME WORDS ON THE LO SHU

OWEN O’SHEA Rope Walk Mount Crozier Cobh County Cork, Ireland e-mail: [email protected]

Last week I received the following e-mail from my imaginary friend, Richard Stein, who is known to his many friends as The Professor. Hi Owen I hope that you are keeping well. I hope that you are behaving yourself in Ireland! I trust that you are still writing the monthly column for The Mathematical Universe. At least I hope that you are! It will help to keep you off the streets, and allow decent people to walk the streets of Cobh in safety! Michelle and I have been doing a little traveling around the good old USA in the last few weeks. We paid a visit to a little town named Odd, which is situated in Raleigh County, West Virginia. It is said that the town of Odd obtained its name in a rather unusual way. A number of people gathered at the local post office with a view to giving a name to their town. Several names were put forward. When one name in particular was suggested, someone in the group responded by saying “That’s odd.” The group then decided to call their town Odd. I do not know if the story is true. I hope it is. In any event, Michelle and I enjoyed out visit to Odd and in general we had a great time on our little break. I met a group of number enthusiasts on our travels. Somehow they knew of me and invited me to give a little talk in the hotel that we were staying in. I normally do 23 © 2009, Baywood Publishing Co., Inc. http://baywood.com

24 / O’SHEA

not do this type of thing when I am on vacation, but I made an exception on this occasion. I gave a little talk on the 3 by 3 magic square, which is, as you probably know, usually referred to by recreational mathematicians as the lo shu. The talk went down very well. Michelle sends her kindest regards. Best wishes, Richard I immediately sent off a reply, asking The Professor to send me details of his talk on the lo shu, saying that I might be able to use the material in my column in The Mathematical Universe. (I ignored his comments about me staying off the streets of Cobh. I realized that The Professor was just being his old self again!) Two days later the Professor sent me the following e-mail. Hi Owen, Hope that you are well. Here are the details of the talk I gave on the lo shu. I hope that you like it, and that you use it in your monthly column in The Mathematical Universe. Heaven knows, you do need some new material to liven up that column! Best regards, Richard New Curiosities on the Lo Shu By Richard Stein

I am here today to say a few words about the lo shu, which is 4 9 2 also known as the 3 by 3 magic square (see Figure 1). I intend to 3 5 7 give mainly new curiosities on the lo shu. I hope that by doing so I will convey to you my belief that the lo shu is one of the most 8 1 6 beautiful mathematical objects known to the human race. A magic square of order n is an arrangement of n2 numbers, Figure 1. usually distinct integers, in the form of a square, such that the The lo shu. sum of the integers in every row, column, and both diagonals equals the same constant. A normal magic square consists of the numbers from 1 to n2. The constant in a normal magic square is equal to (n3 + n)/2. Thus, the constant in the lo shu square is (33 + 3)/2, which equals 15. The lo shu is unique (besides rotations and reflections) in the sense it is the only way to arrange the nine digits so that the sum of each row, each column, and each of the two diagonals is the same. First, note that the first five odd digits placed within the lo shu form a cross, while the four even digits are found in the corners.

NEW CURIOSITIES ON THE LO SHU / 25

Here are some of the basic properties of the lo shu: • 4 + 9 + 2 = 8 + 1 + 6 = 15 • 4 + 3 + 8 = 2 + 7 + 6 = 15 • 42 + 92 + 22 = 82 + 12 + 62 = 101 • 42 + 32 + 82 = 22 + 72 + 62 = 89 • 4 + 5 + 6 = 15 = 2 + 5 + 8 • 42 + 52 + 62 + 22 + 52 + 82 = 170 = 34 · 5 (note the consecutive digits). In addition to the above properties, there are many other interesting and beautiful properties of the lo shu. For example, consider the following two: The middle column, reading down, is 951. Note that 492 – 357 + 816 = 951 and that 294 – 753 + 618 = 159.

Here is a curiosity that I discovered a few years ago: Ignoring the middle column, form two-digit numbers with the other columns as follows: 42 + 37 + 86. These numbers sum to 165. Their sum of their reversals, 68 + 73 + 24, is also 165. The same is true of 84 + 19 + 62 and their reversals, 26 + 91 + 48. Curiously, the sum of the squares of the odd digits, 1, 3, 5, 7, and 9, also equals 165.

In Chapter 21 of his marvelous book Penrose Tiles to Trapdoor Ciphers, Martin Gardner gives a number of remarkable equations involving the numbers that appear in the lo shu. Some of these equations are as follows: • • • •

4922 + 3572 + 8162 = 6182 + 7532 + 2942 4382 + 9512 + 2762 = 6722 + 1592 + 8342 4562 + 9782 + 2312 = 1322 + 8792 + 6542 2582 + 9362 + 4712 = 1742 + 6392 + 8522

The digits in the above equations can be inter-changed, and the equations will still hold. For example, one may have • 2492 + 7352 + 6812 = 8612 + 3752 + 4292 or • 9422 + 5372 + 1862 = 1682 + 5732 + 9242 In the recent past, another remarkable discovery was made. This discovery concerns a relatively unknown problem in mathematics known as the Tarry-Escott problem. The individual who discovered these beautiful equations is unknown to me. The Tarry-Escott problem asks to find two sets of integers, equal in number, such that the integers in each set have the same sum, the same sum of squares, etc., up to and including the same sum of kth powers. Remarkably, the pattern in the lo shu gives a solution to the Tarry-Escott problem. The following beautiful equations hold:

26 / O’SHEA

• 492 + 276 + 618 + 834 = 438 + 816 + 672 + 294 • 4922 + 2762 + 6182 + 8342 = 4382 + 8162 + 6722 + 2942 • 4923 + 2763 + 6183 + 8343 = 4383 + 8163 + 6723 + 2943 Recently I was doing a little doodling on a wet afternoon when I had some free time. I was astonished when I discovered the following equations, which I believe appear here in print for the first time: • 49n + 27n + 61n + 83n = 63n + 81n + 67n + 29n for n = 1, 2, 3. Also, • 92n + 76n + 18n + 34n = 38n + 16n + 72n + 94n for n = 1, 2, 3. Also, • 42n + 26n + 68n + 84n = 48n + 86n + 62n + 24n for n = 1, 2, 3. Also, • 971n + 713n + 139n + 397n = 793n + 931n + 317n + 179n for n = 1, 2, 3. Also, • 97n + 71n + 13n + 39n = 79n + 93n + 31n + 17n for n = 1, 2, 3. I also made the following little discoveries: 1. The sum of the cubes of the first and third columns is 43 + 33 + 83 + 23 + 73 + 63 = 1170 = 234 · 5. Note the consecutive digits. The 2nd, 3rd, and 4th primes are 3, 5, and 7, which appear in the middle row of the lo shu. The sum of the cubes of the 1st and 3rd rows is 43 + 93 + 22 + 83 + 13 + 63 = 1530 = 816 + 2 · 357. 2. The sum of the cubes of the middle column, 93 + 53 + 13 = 855 = (3 · 57) · 5. Note the numbers in the parentheses! 3. The sum of the cubes of the first four even numbers is 43 + 23 + 63 + 83 = 800 = 4 · 268 – 4 – 268. 4. The sum of the cubes of the first five odd numbers is 13 + 33 + 53 + 73 + 93 = 1225 = (1 + 3 + 5 + 7 + 9) (1 – 3 · 5 + 7 · 9). 5. The sum of the squares of the even digits, 2, 4, 6, and 8 equals 120. Curiously, the factorial of the central digit, 5, also equals 120! Also, 120 = 492 – 357 – 8 – 1 – 6. 6. The sum of the squares of the first nine digits is 285. Multiply this by 9 to obtain 2565. The constant of the lo shu is 15. The digits of the middle row are 3, 5, and 7. Note that (3 · 57) · 15 = 2565.

NEW CURIOSITIES ON THE LO SHU / 27

7. The central digit in the lo shu is 5. Note that 9 · (12 + 22 + 32 + 42 + 52) = 495. Curiously, 33 + 53 + 73 also equals 495. 8. The highest number in the lo shu is, of course, 9. The 9th prime is 23. The 23rd prime is 83. Consider the numbers, 3, 5, and 7 across the center of the lo shu. Note that 32 + 52 + 72 = 83. 9. Consider how the digits 4, 3, 8 and 2, 7, 6 are placed in the lo shu. Note that 4 · 38 = 2 · 76. 10. The following two equations are curious: 258 + 456 = 2 · 357 852 + 654 = 2 · 753 11. It is curious that 438 + 9 + 5 + 1 + 276 = 4 · 9 · 2 + 3 · 57 + 81 · 6 = 93 (there are nine cells in the lo shu, and three cells to each side). 12. The four even digits, 2, 4, 6, 8, are used to form the number 2468. Note that 2468 = 492 + 35 · 7 · 8 + 16. The four odd digits, 1, 3, 7, 9, are used to form the number 1379. Note that 1379 = 4 · 38 + 951 + 276. 13. Consider the top digits, 4, 9, 2. The corresponding digits in the bottom row are 8, 1, and 6. Both 49 and 81 are square numbers. The Number 2 is twice the first triangular number and the number 6 is twice the second triangular number. 14. Consider the following equations: 4 · 9 · 2 = 72; 3 · 5 · 7 = 105; 8 · 1 · 6 = 48. Note that 72 – 105 + 48 = 15 (the constant in the lo shu). Also, 72 + 105 + 48 = 225 = 152. 15. Consider the numbers in the top and bottom corners. They are 4, 2, 6 and 8. 42 is twice 21, which is the 8th Fibonacci number; 68 is twice 34, which is the 9th Fibonacci number. There are many other curiosities in the lo shu. For instance, excluding the middle row, consider the numbers in the left and right vertical columns, reading downwards. These are 48 and 26. The number 48 is one less than a square, while 26 is one more than a square. The product of 48 and 26 is 1248, a number consisting of digits in geometrical progression. Also note that 1248 equals 312 · 4. That product contains the first four digits. Also, 1248 = 492 – 3 – 57 + 816. One may also note the following, where the three numbers combined on the right-hand side of the following equations consist of all nine digits: 1248 – 492 = 756 1248 – 357 = 891 1248 – 816 = 432.

28 / O’SHEA

Consider the digits in the top and bottom rows. These are 4, 9, 2 and 8, 1, 6. It is curious that 4 times the 9th prime is 92 and that 8 times the 1st prime is 16. Because the sum of the squares of the digits in the top and bottom rows (and right and left columns) are equal, the following property also holds: Let Tn represent the nth triangular number. Then, T4 + T9 + T2 = T8 + T1 + T6 T4 + T3 + T8 = T2 + T7 + T6. The following equations are curious, because of the appearance of the same digits on the right-hand side of the equations: T42 + T92 + T22 = 2134 T42 + T32 + T82 = 1432 T22 + T72 + T62 = 1234 The numbers 492, 357, and 816 are the legs of Pythagorean right-angle triangles: 4922 + 16452 = 17172; 3572 + 12762 = 13252; and 8162 + 25372 = 26652. The numbers 492, 357, and 816 are also the hypotenuses of Pythagorean right-angle triangles: 1082 + 4802 = 4922; 1682 + 3152 = 3572; and 3842 + 7202 = 8162. One may also note the following curiosities: 492/357 · 816 = 1124 · 57142857 . . . . The nearest integer to this quantity is 1125. Curiously, 1125 = 5 · 152. Also, 1124 = 4 · 92 –3 – 57 + 816. The lo shu has a number of curious connections with 666, the famous number of the beast. The simplest relationship between the two that I know is to consider the digit 4, and the two digits going anti-clockwise underneath it, which are 3 and 9. With these three digits form the number 439. The next three digits going anti-clockwise are 8, 5, and 2. The last three digits going anti-clockwise are 1, 6, and 7. Note that 439 + 8 + 52 + 167 = 666. Alternately, one may take the digits in the lo shu, from left to right and from top to bottom, and form the following sum: 49 + 23 + 578 + 16 = 666. Reversing the order of the digits, the same sum may be obtained as follows: 618 + 7 + 5 + 3 + 29 + 4 = 666. Here is an even more curious relationship between the lo shu and the number 666. Multiple each of the numbers 492, 357, and 816 by 666. One obtains: 492 · 666 = 327672 357 · 666 = 237762 816 · 666 = 543456

NEW CURIOSITIES ON THE LO SHU / 29

The first two results above contain the same digits! Also note that 327672 + 237762 = 565434, a number that contains the same digits as the product of 816 · 666! Consider the three vertical digits (and their reversal) in the middle column of the lo shu: these are 951 and 159. Consider also the three horizontal digits (and their reversal) in the middle row: these are 357 and 753. Now, (9512 – 1592) = 879,120 and (753 2 – 3572) = 439560. Note that 879120 = 2 · 439560. Now consider the numbers (and their reversals) formed by the diagonals: these are 852 and 258 and 654 and 456. One finds that (8522 – 2582) = 659340 and (654 2 – 4562) = 219780. Note that 659340 = 3 · 219780. Using the usual alphabet code, a = 1, b = 2, c = 3, and so on, the sum of the letters in the phrase THE LO SHU MAGIC SQUARE equals 222. Curiously, 222 = 492 – 357 + 81 + 6 = 438 + 9 + 51 – 276. Two final curiosities. I mentioned earlier that 4922 + 3572 + 8162 = 6182 + 7532 + 2942. Both sides of the equation equal 1035369. One finds that 103 · 536 · 9 = 496872. Partition this number as follows: 496: 872. Note that 496 + 872 = 492 + 3 + 57 + 816. It is also curious that 492 + 3 + 57 + 816 = 123 + 456 + 789. BIBLIOGRAPHY M. Gardner, The Return of Dr. Matrix, in Penrose Tiles to Trapdoor Ciphers (Chapter 21), W. H. Freeman and Company, New York, pp. 293-306, 1989. M. Kraitchik, Magic Squares of the Third Order, in Mathematical Recreations, Dover Publications, New York, pp. 146-148, 1953. C. A. Pickover, The Zen of Magic Squares, Circles, and Stars: An Exhibition of Surprising Structures across Dimensions, Princeton University Press, New Jersey, 2002.

J. RECREATIONAL MATHEMATICS, Vol. 35(1) 30-36, 2006

POWER COUPLES

STEVEN KAHAN JOHN W. KENNEDY Mathematics Department Queens College, CUNY Flushing, NY 11367 e-mail: [email protected] or [email protected]

Introduction The following question was recently posed by a curious student of number theory: Is it possible to find positive integers with the property that the sum of their cubes is a perfect square and the sum of their fourth powers is a perfect cube?

A quick check using Mathematica affirms that among the first thousand positive integers, there are precisely two such pairs: (a, b) = (32, 32) and (a, b) = (289, 578). The more interesting issue, and the purpose of this note, is to establish a constructive procedure to generate countably many pairs that satisfy the stated conditions. We accomplish this in two stages, beginning by determining integer pairs that meet the first requirement. Once this has been completed, we employ a similar technique to extract pairs that satisfy the second requirement as well. Construction of Integer Pairs (x, y) for which x3 + y3 is a Perfect Square Choose any two positive integers m and n, m £ n, define u = m3 + n3, and let P(u) represent the prime factorization of u. If every prime in P (u) is raised to an even power, our goal is met, since

30 © 2009, Baywood Publishing Co., Inc. http://baywood.com

POWER COUPLES / 31 a

a

a

a /2

m3 + n 3 = u = p1 1 p 2 2 ... p k k = ( p1 1

a /2

p2 2

a /2 2

... p k k

) ,

which is clearly a perfect square. Otherwise, write P (u) = J(q1q2 . . . ql), where J is a perfect square. Define d = d(u) = q1q2 . . . ql and let x = md and y = nd. Then x3 + y3 = = = =

(md)3 + (nd) 3 (m3 + n3)d 3 ud 3 J(q1q2 . . . q1)(q1q2 . . . ql)3

= J ( q14 q 24 ... q l4 )

= {J 1/ 2 ( q12 q 22 ... q l2 )}2 . That is, x3 + y3 is a perfect square. Note that having obtained an integer pair (x, y) for which x3 + y3 is a perfect square, it is also true that (xc2, yc2) is such a pair for any positive integer c. Examples: 1. Let m = 1 and n = 2. Then u = 13 + 23 = 9 = 32, so that (x, y) = (1, 2) is an integer pair for which x3 + y3 is a perfect square. 2. Let m = 1 and n = 1. Then u = 13 + 13 = 2. Since P (2) = 21, it follows that J = 1 and d = d(2) = 2. Thus, (x, y) = (1 ×2, 1 ×2) = (2, 2) is an integer pair for which x3 + y3 is a perfect square: 23 + 23 = 16 = 42. 3. Let m = 2 and n = 3. Then u = 23 + 33 = 35. Since P (35) = 51 ×71, it follows that J = 1 and d = d(35) = 5 ×7 = 35. Hence, (x, y) = (2 ×35, 3 ×35) = (70, 105) is an integer pair for which x3 + y3 is a perfect square: 703 + 1053 = 343000 + 1157625 = 1500625 = (1225) 2. 4. Let m = 2 and n = 4. Then u = 23 + 43 = 72. Since P (72) = 23 ×32 = (22 ×32) ×2, it follows that J = 22 × 32 = 36 and d = d(72) = 2. Hence, (x, y) = (2 × 2, 4 × 2) = (4, 8) is an integer pair for which x3 + y3 is a perfect square: 43 + 83 = 64 + 512 = 576 = (24)2. Construction of Integer Pairs (a, b) for which a3 + b3 is a Perfect Square and a4 + b4 is a Perfect Cube As discussed in the previous section, let (x, y) be an integer pair for which x3 + y3 is a perfect square. Define v = x4 + y4 and let P (v) denote the prime factorization of v. If every prime in P (v) is raised to a power that is divisible by 3 (that is, each power is congruent to zero mod 3), the problem is solved, since x4 + y4 = v is then a perfect cube and we can choose a = x and b = y.

32 / KAHAN AND KENNEDY

Otherwise, write P (v) = J ( q1 q 2 . . . q k )( r12 r22 ... rl2 ) , where J is a perfect cube. We allow for the possibility that either the set of qs, primes in P (v) that have exponents congruent to 1 (mod 3), or the set of rs, primes in P (v) that have exponents congruent to 2 (mod 3), may be empty. Define d = d(v) = (q1q2 . . . qk)2(r1r2 . . . rl)4. It is obvious that d is a perfect square, so d3 is as well. Moreover, vd = J ( q1 q 2 . . . q k )( r12 r22 ... rl2 )( q1 q 2 ... q k ) 2 ( r1 r2 ... rl ) 4 = J ( q1 q 2 . . . q k ) 3 ( r12 r22 ... rl2 ) 3 = {J 1/ 3 ( q1 q 2 . . . q k )( r12 r22 ... rl2 )}3 is a perfect cube, from which we see that vd4 is as well. Now, let a = xd and b = yd. Then a4 + b4 = (xd)4 + (yd)4 = (x4 + y4)d4 = vd4 is a perfect cube, while a3 + b3 = (xd)3 + (yd)3 = (x3 + y3)d3 is a perfect square, since it is the product of two perfect squares. Observe that if an integer pair (a, b) answers the original question, then for any positive integer c, the pair (ac6, bc6) will also. Examples: 1. Starting with (x, y) = (1, 2), for which x3 + y3 = 9 = 32, let v = 14 + 24 = 17. Since P (17) = 171 and the exponent of 17 is congruent to 1 (mod 3), it follows that J = 1 and d = d(17) = 172 = 289. Then (a, b) = (1 × 289, 2 × 289) = (289, 578) is an integer pair for which a3 + b3 = 2893 + 5783 = 217238121 = (14739) 2, and a4 + b4 = 2894 + 5784 = 118587876497 = (4913) 3 . Consequently, (289c6, 578c6), c = 1, 2, 3, . . . , is a family of solutions. 2. Starting with (x, y) = (2, 2), for which x3 + y3 = 16 = 42, let v = 24 + 24 = 32. Since P (32) = 25 and the exponent of 2 is congruent to 2 (mod 3), it follows that J = 23 = 8 and d = d(32) = 24 = 16. Then (a, b) = (2 × 16, 2 × 16) = (32, 32) is an integer pair for which a3 + b3 = 323 + 323 = 65536 = (256) 2, and a4 + b4 = 324 + 324 = 2097152 = (128) 3 . Again, our previous observation yields (32c6, 32c6), c = 1, 2, 3, . . . , as a family of solutions. 3. Starting with (x, y) = (70, 105), for which x3 + y3 = 12252, let v = 704 + 1054 = 145560625. Since P (145560625) = 54 × 74 × 971 and each of these exponents is

POWER COUPLES / 33

congruent to 1 (mod 3), it follows that J = 53 × 73 = 353 = 42875 and d = d(145560625) = 52 × 72 × 972 = 33952 = 11526025. Then (a, b) = (70 × 33952, 105 × 33952) = (806821750, 1210232625) is an integer pair for which a3 + b3 = 8068217503 + 1210232625 3 = (55 × 75 × 973)2, and a4 + b4 = 8068217504 + 1210232625 4 = (54 × 74 × 973)3 . 4. Starting with (x, y) = (4, 8), for which x3 + y3 = 242, let v = 44 + 84 = 4352. Since P (4352) = 28 × 171, the exponent of 2 is congruent to 2 and the exponent of 17 is congruent to 1 (mod 3), it follows that J = 26 = 43 = 64 and d = d(4352) = 24 × 172 = 682 = 4624. Then (a, b) = (4 × 682, 8 × 682) = (18496, 36992) is an integer pair for which a3 + b3 = 184963 + 369923 = (29 × 31 × 173)2, and a4 + b4 = 184964 + 369924 = (28 × 173)3 . An Extended Problem To add another dimension to the question, we now seek positive integers A and B with the property that the sum of their cubes is a perfect square, the sum of their fourth powers is a perfect cube, and the sum of their sixth powers is a perfect fifth power. Needless to say, we restrict our search, starting only with those integer pairs (a, b) that answer the original question. Let w = a6 + b6 and let P (w) denote the prime factorization of w. If every prime in P (w) is raised to a power that is congruent to 0 (mod 5), the quest is over, since then a6 + b6 = w is a perfect fifth power and we can choose A = a and B = b. Otherwise, write P (w) = J ( q1 q 2 ... )( r12 r22 ... )( s13 s23 ... )( t 14 t 24 ... ), where J is a perfect fifth power, q, r, s, and t are primes whose exponents in P (w) are congruent to 1, 2, 3, and 4 (mod 5), respectively, and some but not all of these sets may be empty. Define d = d(w) = ( q1 q 2 ... ) 24 ( r1 r2 ... )18 ( s1 s2 ... )12 ( t 1 t 2 ... ) 6 ) . Since d is a perfect square, so, too, is d3. Furthermore, since d is a perfect cube, d 4 is also a perfect cube. Finally, wd = J ( q1 q 2 ... )( r12 r22 ... )( s13 s23 ... )( t 14 t 24 ... )( q1 q 2 .. . ) 24 ( r1 r2 ... )18 ( s1 s2 ... )12 ( t 1 t 2 ... ) 6 = J ( q1 q 2 ... ) 25 ( r1 r2 ... ) 20 ( s1 s2 ... )15 ( t 1 t 2 ... )10 = {J 1/ 5 ( q1 q 2 ... ) 5 ( r1 r2 ... ) 4 ( s1 s2 ... ) 3 ( t 1 t 2 ... ) 2 }5

34 / KAHAN AND KENNEDY

is a perfect fifth power, which means that wd 6 is as well. Now, let A = ad and B = bd. Then A6 + B6 = (ad)6 + (bd)6 = (a6 + b6)d 6 = wd 6 is a perfect fifth power, A4 + B4 = (a4 + b4)d4 is a perfect cube, since it is the product of perfect cubes, and A3 + B3 = (a3 + b3)d3 is a perfect square, since it is the product of perfect squares. Furthermore, if (A, B) is an integer pair that meets the aforementioned requirements, then (Ac30, Bc30) is a family of such pairs where c is any positive integer. Examples: 1. Starting with (a, b) = (289, 578), w = 2896 + 5786. Since P(w) = 51 ×131 ×1712, it follows that J = 1710 and d = d(51 × 131 × 1712 ) = 524 × 1324 × 1718. Then (A, B) = (524 × 1324 × 1720, 21 × 524 × 1324 × 1720 ) is an integer pair for which A3 + B3 = 32 × 572 × 1372 × 1760 = (31 × 536 × 1336 × 1730 )2 , A4 + B4 = 596 × 1396 × 1781 = (532 × 1332 × 1727 )3 , and A6 + B6 = 5145 × 13145 × 17120 = (529 × 1329 × 1724 )5 . 2. Starting with (a, b) = (32, 32), w = 326 + 326. Since P(w) = 231, it follows that J = 230 and d = d(231) = 224. Then (A, B) = (229, 229) = (536870912, 536870912) is an integer pair for which A3 + B3 = 288 = (244)2, A4 + B4 = 2117 = (239)3 , and A6 + B6 = 2175 = (235)5 . 3. Starting with (a, b) = (806821750, 1210232625), w = 8068217506 + 1210232625 6. Since P(w) = 518 × 718 × 131 × 611 × 9712, it follows that J = 515 × 715 × 9710 and d = d(518 × 718 × 131 × 611 × 9712) = 512 × 712 × 1324 × 6124 × 9718 . Then (A, B) = (21 × 515 × 715 × 1324 × 6124 × 9720, 31 × 515 × 715 × 1324 × 6124 × 9720) is an integer pair for which A3 + B3 = 546 × 746 × 1372 × 6172 × 9760 = (523 × 723 × 1336 × 6136 × 9730)2,

POWER COUPLES / 35

A4 + B4 = 560 × 760 × 1396 × 6196 × 9781 = (520 × 720 × 1332 × 6132 × 9727 )3, and A6 + B6 = 590 × 790 × 13145 × 61145 × 97120 = (518 × 718 × 1329 × 6129 × 9724 )5 . 4. Starting with (a, b) = (18496, 36992) = (4 ×682, 8 ×682), w = 184966 + 369926. Since P(w) = 236 × 51 × 131 × 1712, it follows that J = 235 × 1710 and d = d(236 × 51 × 131 × 1712) = 224 × 524 × 1324 × 1718 . Then (A, B) = (230 ×524 ×1324 ×1720, 231 ×524 ×1324 ×1720) is an integer pair for which A3 + B3 = 2090 × 32 × 572 × 1372 × 1760 = (245 × 31 × 536 × 1336 × 1730 )2 , A4 + B4 = 2120 × 596 × 1396 × 1781 = (240 × 532 × 1332 × 1727 )3 , and A6 + B6 = 2180 × 5145 × 13145 × 17120 = (236 × 529 × 1329 × 1724 )5 . A similar construction can be employed to go even further. That is if ri ³ 3 , 1£ i £ k , and {r1 , r2 ,... , rk } is a collection of pairwise relatively prime integers, we r +1 r +1 can obtain integer pairs (X, Y) such that X i + Y i is a perfect rith power for each 1 £ i £ k . Related Issues 1. The nonlinear Diophantine equation ar + br = cs, where r ³ 3 and s ³ 2, generalizes the first stage of our constructive procedure in which we treated the case r = 3 and s = 2. Andrew Wiles dispatched this problem quite nicely when r = s in his landmark proof of “Fermat’s Last Theorem.” The method of infinite descent was used by Fermat himself to show that this equation has no solution when r = 4 and s = 2. 2. If a and b are relatively prime, the Diophantine equation ar + br = cs has solutions when r = 3 and s = 2; the pairs (a, b) = (1, 2) and (a, b) = (11, 37) are the two smallest examples. When r = 4 and s = 3, we have verified that this equation has no solution with gcd(a, b) = 1, and we suspect that this is the case for all r ³ 4 and s ³ 3. Removal of the relative primality condition imposed on a and b, however, opens the door to further investigation, as evidenced by the fact that 25 + 25 = 43. 3. To extend the original question in another direction, we can ask if n positive integers exist with the property that the sum of their cubes is a perfect square and the sum of their fourth powers is a perfect cube. We have answered the question in the affirmative when n = 2; the interested reader is invited to pursue it for n > 2. 4. Finally, we mention Beal’s Conjecture, which states that the equation ar + bs = ct has no solution if a, b and c are pairwise relatively prime and r, s, and t are integers greater than or equal to 3. (If we were to allow exponents of 2, many

36 / KAHAN AND KENNEDY

simple solutions exist, among them 13 + 23 = 32, 22 + 112 = 53, and 32 + 24 = 52.) As far as we know, this generalization of Fermat’s Last Theorem has not been proved. Its proposer has offered a sizable monetary prize for either a definitive proof or the production of a counterexample.

J. RECREATIONAL MATHEMATICS, Vo. 35(1) 37-38, 2006

BEHFOROOZ-FRANKLIN MAGIC SQUARE WITH U.S. ELECTION YEARS

HOSSEIN BEHFOROOZ Utica College Utica, New York 13502 e-mail: [email protected]

Years ago, after becoming a U.S. citizen, I got a gift (a coffee mug) with the pictures of all U.S. presidents with their names and dates of election years around a lovely coffee mug. Immediately I thought what the heck, let’s play with these numbers and create a magic square. I used 64 U.S. election years from 1788 to 2040 and created the 8 by 8 magic square with Benjamin Franklin style as shown in Figure 1. This magic square has all Franklin Square properties. Later, just for curiosity, I colored these cells with Blue for Democrats and Red for Republicans and I also put down their party initials (D for Democrats, R for Republicans, F for Federalists, and W for Whigs) in each cell to make it more attractive. I noticed that the right-hand half of the table is filled out with 14 D’s and 18 R’s (no blank cells). The left-hand half has 13 D’s and only 5 R’s and few initials from old parties. There are nine blank cells at the left side for future use. In the middle of the magic square conversation which belongs to recreational mathematics (just for fun), here comes political interests and debates. I am not a fortune teller and I am not looking at this crystal ball to predict something about the future of U.S. elections. But, as we all know, most of the magic squares have symmetric properties, particularly Benjamin Franklin squares. If our Behforooz-Franklin magic square needs to be symmetric, we should witness more R’s (Republicans) in the future. Who said that there are no applications for magic squares?

37 © 2009, Baywood Publishing Co., Inc. http://baywood.com

38 / BEHFOROOZ

Figure 1. Behforooz-Franklin magic square with U.S. election years and magic sum 15312.

J. RECREATIONAL MATHEMATICS, Vol. 35(1) 39-43, 2006

SEMIPENTACUBES

GILBERT LEE University of Victoria ANDY LIU University of Alberta WEN-HSIEN SUN Chiu Chang Mathematics Foundation

Dedicated to the memory of Pieter J. Torbijn Pieter’s first article in the Journal of Recreational Mathematics went back to the second issue (see [1]) and five more followed since then (see [2-6]). He had also been a regular contributor to both the Alphametics and the Problems & Conjectures Sections. A most recent one is the very popular Problem 2621 (see [7]), which is the motivation for this article. By a semipentomino is meant a plane figure consisting of two and a half unit squares joined edge to edge, the half square being a right isosceles triangle. There are only four of them, ad shown in Figure 1 along with their letter codes.

Figure 1. 39 © 2009, Baywood Publishing Co., Inc. http://baywood.com

40 / LEE, LIU AND SUN

In Problem 2621, readers are asked to find all symmetric figures that can be constructed using one, two, three, or four semipantominoes. The numbers of solutions are 1, 5, 9, and 21, respectively. Some of the figures may be constructed in two or more ways. By a semipentacube is meant a solid figure consisting of two and a half unit cubes joined face to face, the half cube being a right isosceles triangular prism. Clearly, a semipentomino raised to the third dimension becomes a semipentacube, but there are two others not obtained this way. They are mirror images of each other, and are shown in Figure 2 along with their letter codes. For our construction problems, we use all six semipentacubes. The number of symmetric figures that are constructible is infinite. We can start with any plane figure constructed from the four semipentominoes and raise them to the third dimension, lying between two parallel planes at unit distance apart. We can then place the remaining two semipentacubes symmetrically on either side of these planes. We have to come up with a logical way of limiting the number of desirable solutions. Such a way comes from the consideration of how the six pieces may be packaged in a box. The total volume of the six semipentacubes is 15, and the smallest box which contains them is 1 × 3 × 5. It is easy to see that no such packing exists. The next smallest boxes are the 1 × 4 × 4 and the 2 × 2 × 4, with empty space of total volume 1. Both packings are possible, but the former admits no symmetric solutions. The latter does admit symmetric solutions, and it is aesthetically more pleasing than the other in any case. If the empty space consists of a unit cube, it has two non-equivalent positions within the 2 × 2 × 2 box. The resulting figures are symmetric, and both are constructible (see Figure 3). When the empty space consists of two half cubes, there are two distinct situations. In the former, each of the half cube is symmetric in respect to itself. In the latter, each of the half cubes is symmetric in respect to the other. There are four possible figures in the former situation, two of which are constructible (see Figure 4). We have determined by exhaustive computer search that the other two have no solutions.

Figure 2.

SEMIPENTACUBES / 41

Figure 3.

Figure 4.

To analyze the latter situation, we have to consider the 16 symmetries for a square-based prism. Four of them involved quarter turns and are irrelevant to our puzzle (see Figure 5). Apart from the identity, the others are: 1. 2. 3. 4. 5. 6. 7.

half turn about an axis joining the centers of opposite square faces; reflections about planes halfway between opposite rectangular faces; reflections about planes halfway between opposite long edges; reflection through the center of the prism; reflection about a plane halfway between opposite square faces; half turns about axes joining the centers of opposite rectangular faces; and half turns about axes joining the midpoints of opposite long edges.

42 / LEE, LIU AND SUN

Figure 5.

A missing half-cube has six feasible positions in a 1 × 2 × 2 layer of the square-based prism. In each of the diagrams given in Figure 6, three layers are shown. The fourth either goes on top or at the bottom. For configuration VI, the fourth layer must go on top. Applying each of the types of symmetries to each of configurations I to VI, we generate all possible figures with symmetries that will fit in 2 × 2 × 4 box. Some figures arise from two symmetry/configuration combinations. Quite a lot of them are not constructible, but there are around 20 with solutions. We leave the investigation to the readers.

SEMIPENTACUBES / 43

Figure 6.

A set of hollow semipentacubes is easy to construct with cardboard paper, even though actual wooden models are not, primarily because of the precision required for the half-cube. After the 28th International Puzzle Party in Prague, this puzzle may become commercially available. It is used by Andy Liu as his Exchange Gift at the IPP. References 1. P. J. Torbijn, Polyiamonds, Journal of Recreational Mathematics, 2:4, pp. 216-277, 1969. 2. P. J. Torbijn, The Unknown World of Octiamonds, Journal of Recreational Mathematics, 7:1, pp. 1-7, 1974. 3. P. J. Torbijn and J. Meeus, Problems with Pentominoes/Heximinoes, Journal of Recreational Mathematics, 10:4, pp. 260-266, 1977-1978. 4. P. J. Torbijn and J. Meeus, Lower Polyominoes on a Checkerboard, Journal of Recreational Mathematics, 26:1, pp. 19-28, 1996. 5. P. J. Torbijn and J. Meeus, Hexicubes, Journal of Recreational Mathematics, 30:2, pp. 89-101, 1999-2000. 6. P. J. Torbijn and J. Meeus, Large Pentomino Rectangles with Central Holes, Journal of Recreational Mathematics, 32:1, pp. 14-24, 2003-2004. 7. P. J. Torbijn, Problem 2621, Journal of Recreational Mathematics, 31:4, p. 300, Solution, 32:4, pp. 342-345, 2003-2004.

SMYOZ: A NEW INDUCTION GAME / 45

1. The ships cannot touch each other (i.e., a ship point belongs to one ship only). 2. When a ship is docked at a shore line, exactly two ship points must be simultaneously also shore points (i.e., ships cannot be docked at any corner and the a-, b-, and g-ships cannot be parallel to the shore line at which the ship is docked). 3. The geometric center of each ship is determined as the center of gravity of the corresponding rectangle. The geometric centers of ships of the same category (b, g, or d) cannot be on the same horizontal or vertical line. 4. If two or more ships are in line, they must be separated from each other by at least two cells. The placement of ships shown in Figure 1a is permissible because the two ships are not in line. In Figure 1b a forbidden placement is shown. 5. If two or more ships are parallel to each other and they are separated by one cell only, they are not allowed to be abreast to the extent of more than one cell. Therefore, the situation of Figure 1c is permissible but that of Figure 1d is not. When all the ships are placed according to these rules, He proceeds further. In each row and column, if two or more ships occupy that row or column, a broken (or red) line must be drawn linking the inner sides of the two outermost ships in that row or column. If there are more than two ships in a given row or column, the broken line will pass through all ships that are in between the two outermost ones (see Figure 2). If there are no ships in a row or column, a dotted (or green) line must be drawn throughout that row or column. In this case the line is called a water line. If there is exactly one ship in a row or column, a dotted (or green) line must be drawn through those cells of the ship that belong to that particular row or column. This line runs in the direction of the given row or column but it does not extend beyond the ship (see Figure 2). Now He declares the number of water lines (w), the number of docked ships (d), and the water point chains ci (i = 1, 2, . . . , n), where ci is the number of water points in the i-th chain and n is the number of chains. Obviously, n

å ci

= 2d + 4

(1)

i=1

These data are listed above the graph (see Figure 2). The value of d can be calculated from Equation (1) if it is not given. These data can be found at the top of Figure 2. He now declares the positions of the b- and g-ships with respect to the position of the a-ship. If a ship is parallel to the a-ship, it is not marked. If it is perpendicular, it is marked with an asterisk (see the left-hand side of Figure 2). He also declares the number of lines inside each ship by 2-digit numbers. The first

46 / HENNYEY AND SZILAGYI

Figure 1.

SMYOZ: A NEW INDUCTION GAME / 47

Figure 2.

SMYOZ: A NEW INDUCTION GAME / 49

Figure 3.

50 / HENNYEY AND SZILAGYI

Figure 4.

SMYOZ: A NEW INDUCTION GAME / 51

Figure 5.

J. RECREATIONAL MATHEMATICS, Vol. 35(1) 52, 2006

LETTER FROM THE EDITOR

This is to announce that the “Ten Year Cumulative Index to the Journal of Recreational Mathematics” is now available for download through the Amazon Kindle program. The URL is http://www.amazon.com/Cumulative-Index-Journal-Recreational-Mathematics/ dp/B001GINP34/ref=sr_1_1?ie=UTF8&s=books&qid=1224859044&sr_1_1 Charles Ashbacher Co-editor, Journal of Recreational Mathematics 5530 Kacena Ave. Marion, IA 52302

52 © 2009, Baywood Publishing Co., Inc. http://baywood.com

J. RECREATIONAL MATHEMATICS, Vol. 35(1) 53-60, 2006

BOOK REVIEWS

Edited by: Charles Ashbacher Charles Ashbacher Technologies Box 294 Hiawatha, IA 52233 e-mail: [email protected] FAX: 928-438-7929 Alternate address: 5530 Kacena Ave. Marion, IA 52302

The Artist and the Mathematician: The Story of Nicholas Bourbaki, the Genius Mathematician Who Never Existed, by Amir D. Aczel, New York: Thunder’s Mouth Press, 2006. 239 pp., $14.95 (paperback). ISBN 1-56858-359-1.

I expect that most JRM readers also know the basic information about Nicholas Bourbaki and his work. Nicholas Bourbaki was the pen name and alter ego of a group of mostly French mathematicians who sponsored seminars and collectively produced a number of comprehensive mathematical texts beginning in the 1930s and continuing throughout the 20th Century. I learned much more about Bourbaki from Aczel’s great little book. Aczel begins with a fascinating and insightful presentation of the personalities of the founding members of Bourbaki, of whom André Weil and Alexandre Grothendieck are most prominent. The very first paragraph of this work briefly describes the puzzling disappearance of Grothendieck in 1991. According to Aczel, Grothendieck suddenly burned 25,000 pages of his own original writings and disappeared into the Pyrenees. Later in Chapter 1, he traces Grothendieck’s early life, in particular his family’s struggle for survival during the Nazi occupation of France. We return to the mysterious Grothendieck in Chapter 14, armed with more facts to support speculation on his motivation and state-of-mind

53 © 2009, Baywood Publishing Co., Inc. http://baywood.com

54 / BOOK REVIEWS

at the time of his disappearance. Aczel calls Grothendieck the most visionary mathematician of the 20th century. Chapter 2 recounts the harrowing experiences of the young André Weil who, like Grothendieck, was caught in the upheaval of World War II. Chapter 3 introduces several other eventual Bourbaki participants including Henri Cartan and Jean Dieudonné. In Chapter 7, we arrive at the founding of the Bourbaki seminar and Weil’s role in inventing the Nicholas Bourbaki persona. Weil was inspired by two wonderful student pranks, which Aczel recounts in Chapter 4. Bourbaki’s main method of mathematical inquiry, which came to be known as structuralism, was adapted from the discipline of linguistics by Claude Lévi-Strauss. Aczel describes structuralism as “a method of intellectual inquiry that provides a framework for organizing and understanding areas of human study concerned with the production and perception of meaning.” This book greatly expanded my appreciation of the significance of Bourbaki through its discussion of structuralism’s influence on existentialism, psychology, psychiatry, and economics. Chapters 14 and 15 cover the decline of Bourbaki. Of prime importance here is Grothendieck’s turn from mathematics to social causes, leading to his eventual disillusionment and disappearance. Weil left France during World War II and thereafter returned only for short visits. Structuralism fell out of favor. Lesser mathematicians replaced founding members of the group. The group no longer publishes and, as Aczel states on page 206, most mathematicians agree that Bourbaki is dead. Ironically, the certitude of this conclusion regarding the fate of the fictional Bourbaki is unmatched by any comparable resolution to the question of the current whereabouts of the enigmatic, but very real, Alexandre Grothendieck. Lamarr Widmer Messiah College Crimes and Mathdemeanors, by Leith Hathout, Wellesley, MA: A. K. Peters Ltd., 2007. 150 pp., $14.96 (paperback). ISBN 978-56881-260-1.

Ravi is a 14-year-old math genius that also specializes in solving mysteries. His tools are mathematics and his ability to reason through problems. There are 14 mysteries in this collection and their names and the mathematics used to solve them are as follows: • A Mystery on Sycamore Lane, the combinatorics of shaking hands; • The Watermelon Swindle, the mathematics of percentages applied to an unknown weight; • An Adventure at the Grand Canyon, geometry and similar angles; • Basketball Intrigue, permutations and combinations;

BOOK REVIEWS / 55

• The Moon Rock, linear combinations of two different weight types used to create specific weights; • A Theft at Dubov Industries, the combinatorics of key combinations; • Murder at the Gambit, probabilities of winning a card game; • A Day at the Race Track, recurrence formulas an derangements; • Bowling Average, sequential trials and lattices; • Caught on Film, the acceleration due to gravity; • A Mishap at Shankar Chemicals, the kinetic energy of liquid flow; • Almost Expelled, the sequential answering of questions about a question until the answer is known; • The Urban Jungle, the mathematics of region covering; and • A Snowy Morning on Oak Street, the rate of snow removal versus the rate of snowfall. Leith Hathout was a high school student when he wrote this book. He shows his love for mathematics in putting forward these problems in the form of crimes to be solved. They are all well written and, with one exception, only high school mathematics is needed to understand the solution. That exception is the last mystery, where calculus is used to describe the rate of snowfall. Mathematics books don’t have to be dull, as many people have demonstrated, it is possible to write exciting stores that use and teach math. Hathout is to be added to the list of people who can make mathematics challenging, interesting, and enjoyable. Charles Ashbacher To Talk of Many Things: An Autobiography, by Dame Kathleen Ollerenshaw, New York: Manchester University Press, 2004. 269 pp., $26.95 (hardbound). ISBN 0-719069874.

Dame Kathleen Ollerenshaw is unusual, even for a mathematician, for she has held two formal careers throughout her long and productive life. She is a gifted mathematician, producing a significant number of quality research papers, and served in political office for nearly three decades. She also managed to accomplish these feats while being almost completely deaf. She was born in 1912, yet did not get her first hearing aid until 1949, four years after receiving her doctorate. It is amazing that she was able to accomplish so much, considering the role of women in society at that time. As she relates, until recently a woman teacher was automatically dismissed when she was to become a mother. Therefore, she also had the third, informal career as a housewife and mother. This book contains very little in the way of mathematics, it is written in the form of a personal diary of her life. Nevertheless, she is obviously a strong person and a role model for female students of mathematics. If I ever teach history of

56 / BOOK REVIEWS

mathematics again, I will strongly recommend her as the subject of a biographical paper or presentation on the part of students. Charles Ashbacher Leonard Euler: A Man to be Reckoned With, by Andreas K. Heyne and Alice K. Heyne, Boston, MA: Birkhauser-Verlag, 2007. 45 pp., $19.95 (hardbound). ISBN 3-7643-8332-1.

A cartoon book about a mathematician seems to be a logical contradiction of the first order. Yet, Euler is most certainly the most prolific mathematician that ever lived and his sponsorship by the rulers of royalist nations involved a great deal of political maneuvering. Wars were fought, there were dynastic changes at the top, and Euler changed residence and sponsors several times, yet he never seemed to miss so much as an epsilon in his mathematical output. The cartoons are humorous, yet still serious enough that a professional mathematician can enjoy the story. Understanding it involves some knowledge of Euler’s life and the other mathematical people that he interacted with, so in that respect younger children will not get some of the humor. Light mathematical reading often appears to be a pointless waste of time for mathematical people. This book is definitely an exception to that maxim. It is a rarity, a cartoon book that could be used as a textbook in the history of mathematics, even at the college level. Younger people will snicker at the captions on page 37, which show a naked Catherine the Great lying on the bed and a disrobing Stanislaw Poniatowski standing by the window. I found Catherine’s strategically placed crown very amusing. Charles Ashbacher Great Feuds in Mathematics, by Hal Hellman, Hoboken, NJ: John Wiley and Sons, 2006. 250 pp., $24.95 (hardbound). ISBN 0-471-64877-9.

Two colleagues saw the cover of this book when I had just started reading it. With some obvious skepticism they asked what kind of feuds there could be in mathematics. Fortunately, at that point I had perused the chapter titles and, as a teacher of the history of mathematics, I was already familiar with the topics of the first three chapters. I was able to give them a brief insight, knowing that as fellow academics they would at least be familiar with priority disputes and plagiarism charges which happen in most any discipline. Each of the ten chapters of this work presents one mathematical conflict. They are in roughly chronological order, allowing for some overlap, beginning with Tartaglia versus Cardano in the 16th century. Chapter 3 covers Newton versus Leibniz, certainly the best-known priority dispute in the history of mathematics

BOOK REVIEWS / 57

and undoubtedly one of the bitterest. The second word in every chapter title is “versus,” with others being Descartes versus Fermat, Bernoulli versus Bernoulli, and Hilbert versus Bouwer. As this brook progresses, the disputes generally become less familiar and the mathematical points of disagreement somewhat more difficult to grasp as the content of our discipline deepens. In addition, some of the later disagreements seem less acrimonious, although it is clear from Hellman’s account that the participants passionately advance their beliefs. This book is well-written and very accessible. The relevant mathematical ideas are clearly presented to the extent needed for understanding the opposing sides of the conflicts. It is balanced, offering more than one plausible interpretation of events or motivations in cases of ambiguity. The focus is on personalities and interaction of the combatants; the pace is lively. After reading this book, you will be able to judge for yourself whether “feud” is too strong a word for the disputes described here. In any case, you will understand that mathematical ideas have indeed been tested and sharpened by competition with opposing viewpoints. That is, these feuds, however nasty, have served a useful purpose. My two colleagues could learn a lot from these accounts and be convinced that we mathematicians are much more lively and passionate than common stereotypes would have one think. Now, if only I can convince them to read Hellman’s book. Lamarr Widmer Messiah College Fallacies in Mathematics, by E. A. Maxwell, New York: Cambridge University Press, 2006. 95 pp., $23.99 (paperback). ISBN 0-521-02640-7.

This re-issue of the classic work first published in 1958 once again demonstrates that it is one of the best short books of mathematics ever published. Each example seems so plausible if you read it quickly, yet all contain a fatal flaw in the reasoning. In many cases that flaw is easy to spot, but in some cases it takes extensive thought before you manage to filter through the grain to the chafe. The fallacies are generally split into topics and the chapter headings are as follows: • • • • • • • • •

The mistake, the howler and the fallacy; Four geometrical fallacies enunciated; Digression on elementary geometry; The “isosceles triangle” fallacy analysed; The other geometrical fallacies analysed; Some fallacies in algebra and trigonometry Fallacies in differentiation; Fallacies in integration; Fallacy by the circular points at infinity;

58 / BOOK REVIEWS

• Some “limit” fallacies; and • Some miscellaneous howlers. After the fallacy is presented with “proof,” a detailed explanation of what has gone wrong is given. This book is an excellent resource for classes on mathematical proof techniques because if you are to do proofs right you need to know how to avoid the pitfalls. Charles Ashbacher A Guide to Complex Variables, by Steven G. Krantz, Washington, DC: The Mathematical Association of America, 2008. 182 pp., $49.95 (hardbound). ISBN 978-0-88385-338-2.

This book is designed to be a review of the principles of complex variables for graduate students and faculty who need an overview. It contains a quick overview of what one ordinarily finds in a course covering complex variables with some brief forays into complex analysis. It is assumed that the person has a high level of mathematical literacy, knowledge of calculus, partial differential equations, and understanding of the basic terms of analysis and topology. A primary focus is to present the fundamental knowledge needed to succeed on a qualifying examination in complex variables. Within this context, the book is unquestionably a success, although the brevity will sometimes be a problem. There are no exercises and examples are sometimes rare. Fortunately, there is an extensive bibliography of additional references. Charles Ashbacher Legacy of the Luoshu: The 4,000 Year Search for the Meaning of the Magic Square of Order Three, by Frank J. Swetz, Wellesley, MA: A. K. Peters, 2008. 228 pp., $35.00 (hardbound). ISBN 978-1-56881-427-8.

The luoshu is arguably the simples of all nontrivial magic squares. It is easy to prove that the smallest nontrivial magic square is the 3 × 3 and the luoshu is formed by using the 4 9 2 smallest positive integers 1 through 9. The magic sum is 15 and the luoshu square is as shown in 3 5 7 Figure 1. According to Chinese legend, this pattern of dots was first found on the back of a turtle, although it seems unlikely that such a pattern is 8 1 6 possible. The Chinese, with their affinity for lore and numerology, incorporated the square into their Figure 1.

BOOK REVIEWS / 59

mysticism, helped by the fact that the single digit numbers held special meaning. In combination with the Chinese belief in the yinyang two-force working of the universe, it became a symbol of harmony, ritual, and fortune-telling to the Chinese. Swetz begins with the Chinese origins and usage of the luoshu and then describes how other cultures also considered it as a talisman and representative of the way the universe works. It is a fascinating story, as it demonstrates how a concept can be used in similar ways in distinct cultures that have no contact with each other. While the modern person generally has grown beyond the idea of “magic numbers,” from this book you can see that historically the “magic” prefix has been an appropriate qualifier for squares of this form. Charles Ashbacher The 15 Puzzle, by Jerry Slocum and Dic Sonneveld, Beverly Hills, CA: Slocum Puzzle Foundation, 2006. 144 pp., $30.00 (hardbound). ISBN 1-890980-15-3.

This is Jerry Slocum at his best, as a historian of puzzles. After a meticulous and colorful account of the early days of the 15 Puzzle, when it took the world by storm and drove it crazy, the authors embark on a painstaking search for the answer to the question: “Who invented the 15 Puzzle”? It reads like detective fiction, with the authors following one trail after another. A central figure in the story is Sam Loyd, a noted American puzzlist who lived around the turn of the 20th century. Although he claimed that he was the inventor of the 15 Puzzle, the authors’ investigation led them to a different conclusion. The most likely inventor was a New York postmaster named Noyes Palmer Chapman, and in computing parlance, the search is not NP-complete. The ultimate irony is that for this exposé of Sam Loyd’s false claim, Slocum won the “Sam Loyd Award.” Next in the book is a mathematical analysis of the 15 Puzzle, by Jerry’s long-time friend and a noted problemist, Dick Hess. (Another very readable proof is given by Sherman Stein in Chapter 14 of his book Mathematics: The Man-Made Universe.) The authors then conclude with some remarks on the ramification of the 15 Puzzle, that it indirectly promotes other sliding-block puzzles. (I should add to that list the puzzle “Rush Hour” by the late Nobiyuki Yoshigahara, another long-time friend of Jerry. Nob is perhaps Japan’s counterpart to Henry Dudeney, just as Jerry is probably contemporary America’s counterpart to Sam Loyd.) Andy Liu University of Alberta

60 / BOOK REVIEWS

Tangram Mater, by Jerry Slocum and Dic Sonneveld, Beverly Hills, CA: Slocum Puzzle Foundation. 206 pp. with four wooden Tangram sets, $17.95. ISBN 1-402705652-6.

This book lays out beautifully 200 Tangram challenges for up to four players. Each puzzle is on a separate page, and the diagram is printed four times facing different directions so that four players can sit around the book and work on the challenges at the same time. The four Tangram sets are made from wood of different colors. At this price, the package is a fantastic bargain. It is a most desirable companion to the author’s definitive treatise on the subject, The Tangram Book. Andy Liu University of Alberta

J. RECREATIONAL MATHEMATICS, Vol. 35(1) 67-71, 2006

PROBLEMS AND CONJECTURES

Edited by: LAMARR WIDMER

Readers are invited to send new problems, solutions, and comments to me at Messiah College, Box 3041, One College Avenue, Grantham, PA 17027 or e-mail to [email protected]. Please put each problem or solution, with your full name and postal address, on a separate sheet. Selection of solutions for publication will take place at least three months after problems appear in print. JRM welcomes original math problems, of a recreational nature, with solutions, if known. The editor may elect to publish a part or an extension of any contributed problem. Previously published problems or variations may be submitted, if they are of particular interest. In such cases, please supply full references. Problems for which no solution is known to the contributor or editor are identified by asterisks (*). Those for which the only known solution requires use of a computer are identified by up-arrows (­). Regular readers of these introductory remarks will know that I happily make use of an undergraduate student assistant in preparation and proofreading of my columns. I am fortunate to teach an ongoing succession of capable mathematics majors whose computer skills exceed my own. I wish to acknowledge the contribution of my latest assistant, Bethany Blackwood, to the Problems and Solutions columns which you will find in this issue. 2750. Fibonacci Search by Pacha Nambi, Shoreline, Washington Find the smallest Fibonacci number which has each of the digits 0 through 9 at least once. 2751. No Small Prime Divisors by Andrew Cusumano, Great Neck, New York The observation that the number N = 2 · 3 · 5 · . . . · pn-1pn + 1 is not divisible by any prime 2, 3, 5, . . . , pn is crucial in Euclid’s well-known proof of the infinitude of the primes. Find other methods of constructing a number which is easily seen to have this property. 67 © 2009, Baywood Publishing Co., Inc. http://baywood.com

68 / PROBLEMS AND CONJECTURES

*2752. Rice Bowl Strategy by Ignotus, Yangan, Myanmar Two players, Cho and Mo Mo, take turns transferring rice grains between a large pile of rice and a bowl which is initially empty. On the first turn, Cho moves one grain from the pile to the bowl and on the second turn, Mo Mo moves two grains from the pile to the bowl. Thereafter, on the nth turn, the players may move n grains from the pile to the bowl or, if the bowl contains at least n grains, she may move n grains from the bowl to the pile. Play continues until the bowl is exactly full or it overflows. The winner is the player who plays last. a. Devise a winning strategy in case the capacity of the bowl is 100 grains. b. Devise a winning strategy for a bowl of capacity N. *2753. Sudoku Grid Count by C. F. Porter, Freeport, Texas a. Consider the 9 × 9 square grid dissected into nine 3 × 3 subsquares. How many different ways are there to arrange the digits 1 through 9 in this grid so that each digit occurs exactly once in every row, column, and subsquare? b. How many of the arrangements in (a) also have the property that each of the two diagonals contains all nine digits? 2754. Mixed Polysticks by Brian Barwell, Hampton, Middlesex, UK Figure 1 shows the polysticks of orders 1 through 4. There are eight polysticks of orders 1 through 3 (Figure 1a) and sixteen of order 4 (the tetrasticks—Figure 1b). Figure 11 of [1] showed that these 24 polysticks can be used to construct the 6 × 6 square grid shown in Figure 2. Find such a solution in which no two of the eight low-order pieces touch one another. Reference:

1. B. R. Barwell, Polysticks, Journal of Recreational Mathematics, 22:3, pp. 165-175, 1990.

2755. MathDice by Richard I. Hess, Rancho Palos Verdes, California For each of the following, you may use addition, subtraction, multiplication, division, powers, decimal points, concatenation, and parentheses, but no roots, factorials, or other notations. Find as many solutions as possible for each part. a. Use each of the digits 2, 3, and 4 exactly once to form an expression that equals 16. b. Use each of the digits 4, 6, and 8 exactly once to form an expression that equals 8. 2756. Postal Problem by Norman Fine, Sewell, New Jersey For a period of time (now past), the first class postage rate in the United States was 37 cents for the first ounce or fraction of an ounce plus 23 cents for each

PROBLEMS AND CONJECTURES / 69

Figure 1.

Figure 2.

70 / PROBLEMS AND CONJECTURES

additional ounce or fraction of an ounce. Aside from the trivial answer of one ounce, what is the smallest package weight which can be exactly paid using only 37-cent stamps? While this question can be answered by an exhaustive search, we desire a more enlightened approach. ­2757. Congruence Relation by Charles Ashbacher, Hiawatha, Iowa On page 14 of [1], Guy notes that Karl Dilcher reports that 31093 º 3 (mod 10932 ) + 1093 + 1). a. Find other values of k such that 3k = 3 (mod k2 + k + 1). b. Find other solutions of bk º b (mod k2 + k + 1). Reference: 1. R. Guy, Unsolved Problems in Number Theory, 3e, Springer-Verlag, New York, 2004.

2758. Expected Length by Daniel Shine, Cincinnati, Ohio A bag initially contains 10 black balls. We repeatedly draw a ball and if it is black, we replace it with a white ball, while if it is white, we return it to the bag. How many repetitions will there be, on average, before all the balls in the bag are white? 2759. Hexomino Square with Holes by Dina Hertzenshtein, Kiriat Bialik, Israel Use one full set of 35 hexominoes, shown in Figure 3, plus an additional copy of the first tile in the second row, to construct the 15 × 15 square with nine empty holes as shown in Figure 4.

Figure 3.

PROBLEMS AND CONJECTURES / 71

Figure 4.

J. RECREATIONAL MATHEMATICS, Vol. 35(1) 90, 2006

PROPOSERS’ AND SOLVERS’ LIST FOR PROBLEMS AND CONJECTURES 34(1)

90 © 2009, Baywood Publishing Co., Inc. http://baywood.com

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JOURNAL OF DRUG EDUCATION Covers psychosocial, pharmacological, legal and social aspects of drugs. Authoritative, timely, adaptable information, insights and methodologies. Serves as a medium for the discussion of all aspects of drug education. JOURNAL OF EDUCATIONAL COMPUTING RESEARCH International, interdisciplinary publication disseminating ideas and research on the applications, effects, and implications of computer-based education to both theorists and practitioners. JOURNAL OF EDUCATIONAL TECHNOLOGY SYSTEMS Reports on the application of technology to the teaching process with emphasis on the use of computers as an integral component of educational systems and teacher-oriented curricula. JOURNAL OF ENVIRONMENTAL SYSTEMS For those concerned with the analysis, design, and management of our environment. Interdisciplinary, incorporating technological and behavioral sciences. JOURNAL OF RECREATIONAL MATHEMATICS Clarifies in a recreational form abstract concepts encountered in formal instruction. Thoughtprovoking, adventurous problems for problem-solving pleasure. Ideal for use in any extra-curricular math activity. JOURNAL OF TECHNICAL WRITING & COMMUNICATION Addressed to technical writers, editors, communication specialists—anyone engaged in the exchange of technical and scientific information. Performs as needed bridge between academia and the world of practitioners. JOURNAL OF WORKPLACE RIGHTS Concerned with any and all human rights, as set forth in the Universal Declaration of Human Rights passed by the U.N. in 1948, that can be affected by the employment relationship. Original theoretical and empirical research. NEW SOLUTIONS: A Journal of Environmental and Occupational Health Policy Explores the growing, changing common ground at the intersection of health, work, and the environment, attempting to both define the issues and offer perspectives for change. NORTH AMERICAN ARCHAEOLOGIST Surveys all aspects of prehistoric and historic archaeology in the United States, Canada, and Northern Mexico within an evolutionary perspective. Emphasis on results of resource management and contract archaeology. OMEGA—JOURNAL OF DEATH AND DYING Provides a psychological study of dying, death, bereavement, suicide, and other lethal behaviors by drawing on contributions from experts in a variety of disciplines and from a wide array of professional settings. THE INTERNATIONAL JOURNAL OF AGING & HUMAN DEVELOPMENT Links theory, research and practice in the exploration of the psychological and social aspects of aging and the aged. Illuminates on the “human” side of gerontology. THE INTERNATIONAL JOURNAL OF PSYCHIATRY IN MEDICINE Fosters research on both psychobiological and psychosocial factors in patient care and the prevention, diagnosis, and treatment of illness. Original research, review articles, novel educational programs, editorials, illustrative case reports.

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