Problem Solving and Recreational Mathematics

Transcription

1 Problem Solving and Recreational Mathematics Paul Yiu Department of Mathematics Florida Atlantic University Summer 016 April 4, 016

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3 Contents 1 Arithmetic Problems Reconstruction of division problems AMM E AMM E AMM E AMM E AMM E Digit problems When can you cancel illegitimately and yet get the correct answer? Repdigits Sums of squares of digits Representation of a number in different bases Base b-representation of a number A number from its base b-representation The Josephus problem Representation of a number in different bases Balanced division by an odd number Balanced base b representation of a number Arithmetic in balanced base 3 representations A matrix card trick Cheney s card trick Three basic principles The pigeonhole principle Arithmetic modulo

4 iv CONTENTS Permutations of three objects Examples A variation: Cheney card trick with spectator choosing secret card The nim game Thenim sum Thenim game Fibonacci and Lucas numbers The Fibonacci sequence Some relations of Fibonacci numbers Zeckendorff representations The Lucas numbers How can we know if a given integer is a Fibonacci number? Periodicity of Fibonacci sequence mod m Counting with Fibonacci numbers Squares and dominos Fat subsets of [n] An arrangement of pennies Counting circular permutations Floor and Ceiling Clock problems Some equations involving x Number of trialing zeros in n! Characterizations of the ceiling and floor functions Combinatorial games Subtraction games The Sprague-Grundy sequence Permutations The universal sequence of permutations The position of a permutation in the universal sequence The disjoint cycle decomposition of a permutation Lexicographical ordering of permutations Dudeney s puzzle

5 CONTENTS v 11.6 Problems on permutations Perfect card shuffles In-shuffles Out-shuffles Perfect shuffling a card to a specified position Interpolation problems What is f(n +1)if f(k) = k for k =0, 1,...,n? What is f(n +1)if f(k) = 1 for k =0, 1,...,n?. 35 k Why is e x not a rational function? Transferrable numbers Right-transferrable numbers Left-transferrable integers The equilateral lattice L (n) Counting triangles Counting parallelograms Counting regular hexagons Counting triangles Integer triangles of sidelengths n Integer scalene triangles with sidelengths n Number of integer triangles with perimeter n The partition number p 3 (n) Pythagorean triangles Primitive Pythagorean triples Rational angles Some basic properties of primitive Pythagorean triples A Pythagorean triangle with an inscribed square When are x px ± q both factorable? Dissection of a square into Pythagorean triangles The area of a triangle Heron s formula for the area of a triangle Heron triangles The perimeter of a Heron triangle is even

6 vi CONTENTS 18.. The area of a Heron triangle is divisible by Heron triangles with sides < Heron triangle as lattice triangle Indecomposable Heron triangles Heron triangles with area equal to perimeter Heron triangles with integer inradii Heron triangles with square areas The golden ratio Division of a segment in the golden ratio The regular pentagon Construction of 36, 54, and 7 angles The most non-isosceles triangle The Arbelos Archimedes twin circle theorem Incircle of the arbelos Construction of incircle of arbelos Archimedean circles in the arbelos Menelaus and Ceva theorems Menelaus theorem Ceva s theorem Routh and Ceva theorems Barycentric coordinates Cevian and traces Area and barycentric coordinates Perimeter - area bisectors of a triangle Construction of perimeter-area bisectors in angle A Isosceles triangles Scalene triangles Triangles with exactly two perimeter-area bisectors Integer triangles Triangles with D a < Triangles with D a = Triangles with D a > Very Small Integer Triangles 357

7 CONTENTS vii 5 The games of Euclid and Wythoff The game of Euclid Wythoff s game Beatty s Theorem Number sequences The Farey sequence The number spiral The prime number spirals The prime number spiral beginning with The prime number spiral beginning with Two enumerations of the rational numbers in (0,1) Inequalities The inequality AGH Proofs of the A-G inequality The AGH mean inequality Cauchy-Schwarz inequality Convexity The power mean inequality Bernoulli inequality Holland s Inequality Miscellaneous inequalities A useful variant of AG Calculus problems Philo s line Triangle with minimum perimeter Maxima and minima without calculus Quartic polynomials and the golden ratio Infinite Series Power series The Bernoulli numbers and the tangent function The Euler numbers and the secant function Summation of finite series

8 viii CONTENTS 31 The art of integration Indefinite integrals An iterated integral and a generalization Crux Math A generalization Rearrangements of the alternating harmonic series Riemann s theorem on conditionally convergent series Rearrangements of the alternating harmonic series

9 Chapter 1 Arithmetic Problems 1.1 Reconstruction of division problems AMM E1 This is Problem E1 of the AMERICAN MATHEMATICAL MONTHLY: x 7 x x x x x x) x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Clearly, the last second digit of the quotient is 0. Let the divisor be the 3-digit number d. Consider the 3-digit number in the seventh line, which is a multiple of d. Its difference from the 4-digit number in the sixth line is a -digit number. This must be 9xx. This cannot be the same as the 3-digit number in the fifth line, since the difference between the 3-digit numbers in the fourth and fifth lines is a 3-digit number. Therefore, in the quotient, the digit after 7 is a larger one, which must be smaller than the first and the last digits, since these give 4-digit multiples of d. It follows that the quotient is

10 10 Arithmetic Problems Since 8d is a 3-digit number 9xx, the 4-digit number in the third and bottom lines is 9d =10xx or 11xx. From this 8d must be 99x, and therefore 99 = )

11 1.1 Reconstruction of division problems AMM E10 This is Problem E10 of the MONTHLY, by Fitch Cheney. In this case, not even one single digit is given. x x x x x x x x x) x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

12 104 Arithmetic Problems AMM E1111 This is said to be the most popular MONTHLY problem. It appeared in the April issue of Our good friend and eminent numerologist, Professor Euclide Paracelso Bombasto Umbugio, has been busily engaged testing on his desk calculator the possible solutions to the problem of reconstructing the following exact long division in which the digits indiscriminately were each replaced by x save in the quotient where they were almost entirely omitted. x x 8 x x x x x) x x x x x x x x x x x x x x x x x x x x x x x x x x Deflate the Professor! That is, reduce the possibilities to ( ) 0. Martin Gardner s remark: Because any number raised to the power of zero is one, the reader s task is to discover the unique reconstruction of the problem. The 8 is in correct position above the line, making it the third digit of a five-digit answer. The problem is easier than it looks, yielding readily to a few elementary insights.

13 1.1 Reconstruction of division problems AMM E971 Reconstruct the division problem ) Charles Twigg comments that if the digit is replaced by 9, the answer is also unique. ) 9

14 106 Arithmetic Problems AMM E198 Here is a multiplication problem: A multiplication of a three-digit number by -digit number has the form in which all digits involved are prime numbers. Reconstruct the multiplication. (Note that 1 is not a prime number). p p p p p p p p p p p p p p p p p p

15 Chapter Digit problems.1 When can you cancel illegitimately and yet get the correct answer? Let ab and bc be -digit numbers. When do such illegitimate cancellations as ab bc = a b bc = a c, allowing perhaps further simplifications of a? c Answer. 16 = 1, 19 = 1, 6 =, 49 = Solution. We may assume a, b, c not all equal. Suppose a, b, c are positive integers 9 such that 10a+b = a. 10b+c c (10a + b)c = a(10b + c),or(9a + b)c =10ab. If any two of a, b, c are equal, then all three are equal. We shall therefore assume a, b, c all distinct. 9ac = b(10a c). If b is not divisible by 3, then 9 divides 10a c =9a +(a c). It follows that a = c, a case we need not consider. It remains to consider b =3, 6, 9. Rewriting (*) as (9a + b)c =10ab. If c is divisible by 5, it must be 5, and we have 9a + b =ab. The only possibilities are (b, a) =(6, ), (9, 1), giving distinct (a, b, c) =(1, 9, 5), (, 6, 5). If c is not divisible by 5, then 9a + b is divisible by 5. The only possibilities of distinct (a, b) are (b, a) =(3, 8), (6, 1), (9, 4). Only the

16 108 Digit problems latter two yield (a, b, c) =(1, 6, 4), (4, 9, 8). Exercise 1. Find all possibilities of illegitimate cancellations of each of the following types, leading to correct results, allowing perhaps further simplifications. (a) a bc = c, b ad d (b) c a b = c, d b a d (c) a b c = a. b cd d. Find all 4-digit numbers like 1805 = 19 5, which, when divided by the its last two digits, gives the square of the number one more than its first two digits.

17 . Repdigits 109. Repdigits A repdigit is a number whose decimal representation consists of a repetition of the same decimal digit. Let a be an integer between 0 and 9. For a positive integer n, the repdigit a n consists of a string of n digits each equal to a. Thus, a n = a 9 (10n 1). Example.1. Show that 16 n 6 n 4 = 1 4, 19 n 9 n 5 = 1 5, 6 n 6 n 5 = 5, 49 n 9 n 8 = 4 8. Solution. More generally, we seek equalities of the form abn = a for b nc c distinct integer digits a, b, c. Here, ab n is digit a followed by n digits each equal to b. To avoid confusion, we shall indicate multiplication with the sign. The condition (ab n ) c =(b n c) a is equivalent to ( 10 n a + b ) ( ) 10b 9 (10n 1) c = 9 (10n 1) + c a, ( (10 n 1)a + b ) ( ) 10b 9 (10n 1) c = 9 (10n 1) a. Cancelling a common divisor 10n 1, we obtain (9a + b)c =10ab, which 9 is the same condition for ab = a. bc c Exercise 1. Complete the following multiplication table of repdigits. 1 n n 3 n 4 n 5 n 6 n 7 n 8 n 9 n 1 1 n n 3 n 4 n 5 n 6 n 7 n 8 n 9 n 4 n 6 n 8 n 1 n0 13 n 1 15 n n n n 13 n 1 16 n n n n n n 1 6 n0 6 n n n 1 39 n n n0 38 n n0 49 n n n 1 53 n 8 59 n n 39 6 n n n n n 01. Simplify (1 n )(10 n 1 5). Answer. (1 n )(10 n 1 5) = 1 n 5 n.

18 110 Digit problems Exercise 1. Find the three 3-digit numbers each of which is equal to the product of the sum of its digits by the sum of the squares of its digits. Answer. 133, 315, Find all 4-digit numbers abcd such that 3 abcd = a + b + c + d. Answer and 583. Solution. There are only twelve 4-digit numbers which are cubes. For only two of them is the cube root equal to the sum of digits. n n Use each digit 1,, 3, 4, 5, 6, 7, 8, 9 exactly once to form prime numbers whose sum is smallest possible. What if we also include the digit 0? 4. There are exactly four 3-digit numbers each equal to the sum of the cubes of its own digits. Three of them are 153, 371, and 407. What is the remaining one? 5. Find all possibilities of a 3-digit number such that the three numbers obtained by cyclic permutations of its digits are in arithmetic progression. Answer. 148, 185, 59, 96. Solution. Let abc be one such 3-digit numbers, with a smallest among the digits (which are not all equal). The other two numbers are bca and cab. Their sum abc + bca + cab = 111 (a + b + c). Therefore the middle number = 37 (a + b + c). We need therefore look for numbers of the form abc =37 k with digit sum equal to s, and check if 37 s = bca or cab. We may ignore multiples of 3 for k (giving repdigits for 37 k). Note that 3k < 7. We need only consider k =4, 5, 7, 8. k 37 k s 37 s arithmetic progression = , 481, = , 518, = 59 59, 59, = 69 96, 69, 96

19 . Repdigits A 10-digit number is called pandigital if it contains each of the digits 0, 1,..., 9 exactly once. For example, is pandigital. We regard a 9-digit number containing each of 1,...,9 exactly once as pandigital (with 0 as the leftmost digit). In particular, the number A := is pandigital. There are exactly 33 positive integers n for which na are pandigital as shown below. n na n na n na How would you characterize these values of n? 7. Find the smallest natural number N, such that, in the decimal notation, N and N together use all the ten digits 0, 1,...,9. Answer. N = and N = 6970.

20 11 Digit problems.3 Sums of squares of digits Given a number N = a 1 a a n of n decimal digits, consider the sum of digits function For example s(n) =a 1 + a + + a n. s(11) =, s(56) = 41, s(85) = 89, s(99) = 16. For a positive integer N, consider the sequence S(N) : N, s(n), s (N),...,s k (N),..., where s k (N) is obtained from N by k applications of s. Theorem.1. For every positive integer N, the sequence S(N) is either eventually constant at 1 or periodic. The period has length 8 and form a cycle A proof of the theorem is outlined in the following exercise.

21 .3 Sums of squares of digits 113 Exercise 1. Prove by mathematical induction that 10 n 1 > 81n for n 4.. Prove that if N has 4 or more digits, then s(n) <N. Solution. If N has n digits, then (i) N 10 n 1, (ii) s(n) 81n. From the previous exercise, for n 4, N 10 n 1 > 81n s(n). 3. Verify a stronger result: if N has 3 digits, then s(n) <N. Solution. We seek all 3-digit numbers N = abc for which s(n) N. (i) Since s(n) 43, we need only consider n 43. (ii) Now if a =, then s(n) =4+b + c 186 <N. Therefore a =1. (iii) s(n) =1 + b + c is a 3-digit number if and only if b + c 99. Here are the only possibilities: (b, c) (5, 9) (6, 9) (7, 9) (8, 9) (9, 9) (6, 8) (7, 8) (8, 8) (9, 5) (9, 6) (9, 7) (9, 8) (8, 6) (8, 7) s(n) Therefore there is no 3-digit number N satisfying s(n) N. 4. For a given integer N, there is k for which s k (N) is a -digit number. 5. If n is one of the integers 1, 7, 10, 13, 19, 3, 8, 31, 3, 44, 49, 68, 70, 79, 8, 86, 91, 94, 97, then s k (N) =1for some k

22 114 Digit problems 6. If N is a -digit integer other than 1, 7, 10, 13, 19, 3, 8, 31, 3, 44, 49, 68, 70, 79, 8, 86, 91, 94, 97, then the sequence S(N) is eventually cycling between 4, 16, 37, 58, 89, 145, 4,

23 Chapter 3 Representation of a number in different bases 3.1 Base b-representation of a number Let b be a fixed positive integer. To write an integer n in base b, keep on dividing the number by b until the quotient is 0, and record the remainders, which are integers between 0 and b 1, from right to left. The resulting sequence, which always begins with a nonzero leftmost digit, is the representation of n in base b. Example 3.1. (a) 13 = [ ] divisor quotient remainder

24 116 Representation of a number in different bases (b) 13 = [1110] 3 divisor quotient remainder A number from its base b-representation Given the base b representation of an integer: n =[a 0 a 1 a k 1 a k ] b, to find the integer in its decimal form, calculate a sequence of numbers n 0, n 1,...,n k as follows: (i) n 0 = a 0, (ii) for i =1,,...,k, n i = a i + b n i 1. Then, n = n k. Example 3.. (a) [ ] = 13. (b) [3147] 11 = Example 3.3. Ask your friend to think of a number between 1 and 31. Then ask her if the number appears in card (a), (b), (c), (d), or (e). Now add up the numbers from the lower left hand corners of the card which she answers yes. That is the number she has chosen.

25 3.1 Base b-representation of a number

26 118 Representation of a number in different bases Exercise 1. Complete the multiplication table in base 7.. Multiply in base 7: [1346] 7 [06] 7 = [1346] 7 [15] 7 = [1346] 7 [4] 7 = [1346] 7 [33] 7 = [1346] 7 [4] 7 = [1346] 7 [51] 7 = 3. Ask your friend to write down a polynomial f(x) with nonnegative integer coefficients. Ask her for the value of f(1). She returns 7. Ask her for the value of f(8). She returns What is the polynomial?

27 3. The Josephus problem The Josephus problem There are n people, numbered consecutively, standing in a circle. First, Number sits down, then Number 4, Number 6, etc., continuing around the circle with every other standing person sitting down until just one person is left standing. What number is this person? This is Problem 1031 of MATHEMATICS MAGAZINE, a reformulation of the Josephus problem. Here is M. Chamberlain s solution: Write n = m + k where 0 k m 1. Then seat the people numbered, 4,...,k. This leaves m people standing, beginning with the person numbered k +1; call him Stan. Now continue to seat people until you get back to Stan. It is easy to see that m 1 people will be left standing, starting with Stan again. On every subsequent pass of the circle half of those standing will be left standing with Stan always the first among them. Stan s the man. 6 * Let J(n) be the position of the last standing man in a circle of n. Theorem 3.1. The binary expansion of J(n) is obtained by transferring the leftmost digit 1 of the binary expansion of n to the rightmost. Proof. If n = m + k for k m, then J(n) =(n m )+1. The binary expansion of J(n) is obtained by moving the leftmost digit of the binary expansion of n (corresponding to m ) to the rightmost. 10 5

28 10 Representation of a number in different bases Exercise 1. For what values of n is J(n) =n?. For what values of n is J(n) =n 1?

29 Chapter 4 Representation of a number in different bases 4.1 Balanced division by an odd number Given a positive integer b and a string of b consecutive integers, for every integer a, there are unique integers q and r so that a = bq + r with r in the given string of b consecutive integers. In particular, if b is odd, say b =c +1, we may choose the string of remainders to be c, (c 1),..., 1, 0, 1,..., c 1, c. If we write a = bq + r, c r c, we say that this is the balanced division of a by the odd number b. 4. Balanced base b representation of a number Let b be a positive odd number. The balanced base b representation of an integer n is obtained by repeatedly performing balanced divisions by b and recording, from right to left, the remainders, which are integers in the range b 1 b 1,, 1, 0, 1,..., 1,, 3, 4,.... We shall write in place of the negative remainders 1,, 3, 4,...

30 1 Representation of a number in different bases Examples (1) 100 = divisor quotient remainder The balanced ternary form of a number expresses it as a sum and/or differences of distinct powers of 3: 100 = Here is a simple application of the balanced ternary expansion of numbers. Example 4.1. Suppose we have a beam balance and a set of seven standard weights 1,, 4, 8, 16, 3, 64 units. It is possible to weigh every integer units from 1 to 17 with these seven standard weights. For example, since 100 = , balance is achieved by putting a weight of 100 units on one pan, and standard weights of 4, 3 and 64 units on the other. However, with a set of five standard weights 1, 3, 9, 7, 81, itis possible to weigh every integer units from 1 to 11. For example, since =11101, a weight of 100 units can be balanced by putting it with a standard weight 9 units on one pan and the standard weights 1, 7, 81 on the other pan.

31 4.3 Arithmetic in balanced base 3 representations Arithmetic in balanced base 3 representations Example =

32 14 Representation of a number in different bases A matrix card trick Let p and q be given odd numbers, say p =h +1and q =k +1. Arrange pq cards in the form of a p q matrix. Ask a spectator to select one secret card and name the column containing the secret card. Then the cards are (i) picked up along columns in any order, except that the named one is the middle one, being preceded by k arbitrary columns and followed by the remaining k columns arbitrarily, (ii) rearranged into the p q matrix along rows. Repeat a number of times, each time asking for the column which contains the secret card. The secret card will eventually appear exactly in the middle row of the matrix. If p q, you can tell what the secret card is after the first step. It is in the middle row. We shall henceforth assume p>qso that there are more rows than columns. Example 4.3. In the arrangement below, the spectator chooses and tells you that it is in the second column J Q K 9 A Then you pick up, in order, the first, the second, and then the third columns, and deal the cards in rows: 5 J Q A 6 8 K Now ask the spectator which column contains the secret card. She will answer the first column. Then you pick up, in order, the third, the first, and the second column, and deal the cards in rows:

33 4.3 Arithmetic in balanced base 3 representations K Q 6 J 10 A 8 10 Now, the spectator will tell you that the secret card is in the third column. You immediately know that the card is. How many times do you need to rearrange the cards before you can tell what the secret card is? 1 Label the rows h, (h 1),...,0,...,h 1, h so that the middle rowisrow0. An application of (i) and (ii) will bring the i-th card in the column containing the secret card into row g(i), where g(i) is the quotient in the balanced division of i by q. Equivalently, g(i) =b if i = bq + r for k i k. It is easy to see that g is an odd function: under the same condition, i = bq r, with k r k, so that g( i) = b. Now, g(i) is decreasing for positive i, and increasing for i<0. Furthermore, if g(i) =0, then all subsequent iterates by g are stationary at 0. Therefore, if we write p =h +1, then the number of times to guarantee the secret card in the middle row is min{l : g l (h) =0}. The function g on positive integers can be easily described in terms of the base q representations of numbers. Lemma 4.1. If n = a m a m 1...a 1 a 0 q in balanced base q representation, then g(n) = a m a m 1 a 1 q. Proposition 4.. The number of times for the secret card to move to the middle row is not more than the length of the balanced base q- representation of the integer p 1. Examples (1) 51 cards in 3 columns: p =17, q =3. Since 8= 101 3, three operations suffice. 1 G. R. Sanchis, A card trick and the mathematics behind it, College Math. Journal, 37 (006) Here, we make use of balanced base q representation of numbers.

34 16 Representation of a number in different bases () 35 cards in 5 columns: p =7, q =5. Since 3= 1 5,two operations suffice. Exercise Find the number of operations required to move the secret card to the middle row for cards in 5 columns;. 39 cards in three columns.

35 Chapter 5 Cheney s card trick You are the magician s assistant. What he will do is to ask a spectator to give you any 5 cards from a deck of 5. You show him 4 of the cards, and in no time, he will tell everybody what the 5th card is. This of course depends on you, and you have to do things properly by making use of the following three basic principles. 5.1 Three basic principles The pigeonhole principle Among 5 cards at least must be of the same suit. So you and the magician agree that the secret card has the same suit as the first card Arithmetic modulo 13 The distance of two points on a 13-hour clock is no more than 6. We decide which of the two cards to be shown as the first, and which to be kept secret. For calculations, we treat A, J, Q, and K are respectively 1, 11, 1, and 13 respectively. Now you can determine the distance between these two cards. From one of these, going clockwise, you get to the other by travelling this distance on the 13-hour clock. Keep the latter as the secret card.

36 18 Cheney s card trick hours distance clockwise and 7 5 to 7 3 and to 3 and J 4 J to A and to A The following diagram shows that the distance between 3 and 10 is 6, not 7. Q K A J Permutations of three objects There are 6 arrangements of three objects The remaining three cards can be ordered as small, medium, and large. 1 Now rearrange them properly to tell the magician what number he should add (clockwise) to the first card to get the number on the secret card. Let s agree on this: arrangement distance sml 1 slm msl 3 mls 4 lsm 5 lms 6 If you, the assistant, want to tell the magician that he should add 4 to the number (clockwise) on the first card, deal the medium as the second card, the large as the third, and the small as the fourth card. 1 First by numerical order; for cards with the same number, order by suits: < < <.

37 5. Examples Examples Example 5.1. Suppose you have 5, 7, J, 4, and Q, and you decide to use the cards for the first and the secret ones. The distance between 7 and 4 is of course 3, clockwise from 4 to 7. You therefore show 4 as the first card, and arrange the other three cards, 5, J, and Q, in the order medium, small, large. The second card is J, the third 5, and the fourth Q. The secret card is 7. 4 J 5 Q?? Example 5.. Now to the magician. Suppose your assistant show you these four cards in order: Q J 7 4?? Then you know that the secret card is a, and you get the number by adding to Q the number determined by the order large, medium, small, which is 6. Going clockwise, this is 5. The secret card is 5. Exercise 1. For the assistant: (a) 5, 7, 6, 5, Q. (b), J, K,, 8.

38 130 Cheney s card trick. For the magician: what is the secret card? J 8????

39 5.3 A variation: Cheney card trick with spectator choosing secret card A variation: Cheney card trick with spectator choosing secret card Now the spectators say they would choose the secret card. What should you, the assistant, do? 1. Arrange the four open cards in ascending order, and put the n-th card in the first position, with n =1,, 3, 4, according as the secret card is for,,, or. For example, if you are given, 6, K, 10, and 4 as secret card, then since you put in the first position. < 6 < 10 < K,. Note the rank of the secret card. If it is the same as the first card, put the secret card in the third position, and arrange the remaining three in any order. 3. If the ranks of secret card and the first card are different, determine the 13 point clock difference, and arrange the remaining three open cards to indicate this difference. (a) If the rank of the secret card is higher than that of the first card, put the secret card in the second position, followed by the remaining three open cards in the positions indicating the difference. (b) If the rank of the secret card is lower than that of the first card, put the three open cards in the positions indicating the difference, followed by the secret card.?? 6 K 10 Now, for the magician,

40 13 Cheney s card trick 1. Determine the order of the first card among the four open cards. The secret card is,,, according as this is the least, second, third, or largest.. If the secret card appears in the third position, its rank is the same as the rank of the first card. 3. If the secret card separates the first card from the remaining three, add the number determined by the remaining three cards to the first card to get the rank of the secret card. 4. If the secret card appears at the end, subtract the number determined by the remaining three cards from the first card to get the rank of the secret card. Exercise 1. For the assistant: (a) 5, 7, 6 (secret card), 5, Q. (b), J, K,, 8 (secret card).. For the magician: what is the secret card? 5?? J 8??

41 Chapter 6 The nim game 6.1 The nim sum The nim sum of two nonnegative integers is the addition in their base notations without carries. If we write 0 0=0, 0 1=1 0=1, 1 1=0, then in terms of the base expansions of a and b, a b =(a 1 a a n ) (b 1 b b n )=(a 1 b 1 )(a b ) (a n b n ). The nim sum is associative, commutative, and has 0 as identity element. In particular, a a =0for every natural number a. Example 6.1. (a) 6 5 = [110] [101] = [011] =3. (b) = [100010] [111001] = [011011] =7. Here are the nim sums of numbers 15:

42 134 The nim game The nim game Given three piles of marbles, with a, b, c marbles respectively, players A and B alternately remove a positive amount of marbles from any pile. The player who makes the last move wins. Theorem 6.1. In the nim game, the player who can balance the nim sum equation has a winning strategy. Therefore, provided that the initial position (a, b, c) does not satisfy a b c =0, the first player has a winning strategy. For example, suppose the initial position is (1, 7, 9). Since 1 9 = 5, the first player can remove marbles from the second pile to maintain a balance of the nim sum equation 1 5 9=0, thereby securing a winning position. Remarks. (1) This theorem indeed generalizes to an arbitrary number of piles. () The Missère Nim game: Suppose now we change the rule: in the Nim game, the player who makes the last move loses. Here is a winning strategy: Play as for ordinary Nim, until you can move to a position in which all piles have just one marble.

43 6. The nim game 135 Exercise In each of the following nime games, it is your turn to move. How would you ensure a winning position? (a) 3, 5, 7 marbles. (b) 9, 10, 1 marbles. (c) 1, 8, 9 marbles. (d) 1, 10, 1 marbles.

44 136 The nim game

45 Chapter 7 Fibonacci and Lucas numbers 7.1 The Fibonacci sequence The Fibonacci numbers F n are defined recursively by F n+1 = F n + F n 1, F 0 =0,F 1 =1. The first few Fibonacci numbers are n F n An explicit expression can be obtained for the Fibonacci numbers by finding their generating function, which is the formal power series F (x) :=F 0 + F 1 x + + F n x n +. From the defining relations, we have F x = F 1 x x + F 0 x, F 3 x 3 = F x x + F 1 x x, F 4 x 4 = F 3 x 3 x + F x x,. F n x n = F n 1 x n 1 x + F n x n x,.

46 138 Fibonacci and Lucas numbers Combining these relations we have F (x) (F 0 + F 1 x)= (F (x) F 0 ) x + F (x) x, F (x) x = F (x) x + F (x) x, (1 x x )F (x) = x. Thus, we obtain the generating function of the Fibonacci numbers: x F (x) = 1 x x. There is a factorization of 1 x x by making use of the roots of the quadratic polynomial. Let α>βbe the two roots. We have α + β =1, αβ = 1. More explicitly, α =, β =. Now, since 1 x x =(1 αx)(1 βx),wehaveapartial fraction decomposition x 1 x x = x (1 αx)(1 βx) = 1 ( 1 α β 1 αx 1 ). 1 βx 1 Each of and 1 has a simple power series expansion. In fact, 1 αx 1 βx making use of 1 1 x =1+x + x + + x n + = and noting that α β = 5,wehave ( x 1 x x = 1 α n x n 5 = n=0 n=0 α n β n 5 x n. x n, n=0 ) β n x n n=0 The coefficients of this power series are the Fibonacci numbers: F n = αn β n 5, n =0, 1,,...

47 7.1 The Fibonacci sequence F n is the integer nearest to αn 5 : F n = α n 5. Proof. F n αn 5 = β n < 1 < For n, F n+1 = αf n. Proof. Note that F n+1 αf n = αβn β n+1 5 = α β 5 β n = β n. For n, F n+1 αf n = β n < lim n F n+1 F n = α. Exercise 1. (a) Make use of only the fact that 987 is a Fibonacci number to confirm that is also a Fibonacci number, and find all intermediate Fibonacci numbers. (b) Make use of the result of (a) to decide if 7506 is a Fibonacci number.. Prove by mathematical induction the Cassini formula: F n+1 F n 1 F n =( 1) n. 3. The conversion from miles into kilometers can be neatly expressed by the Fibonacci numbers. miles kilometers How far does this go? Taking 1 meter as inches, what is the largest n for which F n miles can be approximated by F n+1 kilometers, correct to the nearest whole number?

48 140 Fibonacci and Lucas numbers 4. Prove the Fermat Last Theorem for Fibonacci numbers: there is no solution of x n + y n = z n, n, in which x, y, z are (nonzero) Fibonacci numbers. 7. Some relations of Fibonacci numbers 1. Sum of consecutive Fibonacci numbers: n F k = F n+ 1. k=1. Sum of consecutive odd Fibonacci numbers: n F k 1 = F n. k=1 3. Sum of consecutive even Fibonacci numbers: n F k = F n+1 1. k=1 4. Sum of squares of consecutive Fibonacci numbers: n Fk = F n F n+1. k=1 5. Cassini s formula: F n+1 F n 1 Fn =( 1) n.

49 7.3 Zeckendorff representations Zeckendorff representations Every positive integer can be written uniquely as a sum of distinct Fibonacci numbers with no consecutive indices: N = F k1 + F k + + F km,k i >k i+1 +1,i<m. This is called the Zeckendorff representation of N. How do you find the Zeckendorff representation of , knowing that the largest Fibonacci summand is F 30 = Exercise 1. Prove that F 1 + F F k 1 =F k 1, F + F F k =??. (a) Suppose Fibonacci had wanted to set up an annuity that would pay F n lira on the nth year after the plan was established, for n = 1,, 3,... (F 1 = F = 1,F n = F n 1 + F n for n > ). To fund such an annuity, Fibonacci had planned to invest a sum of money with the Bank of Pisa, which paid 70% interest per year, and instruct them to administer the trust. How much money did he have to invest so that the annuity could last in perpetuity? (b) When he got to the bank, Fibonacci found that their interest rate was only 7% (he had misread their ads), not enough for his purposes. Despondently, he went looking for another bank with a higher interest rate. What rate must he seek in order to allow for a perpetual annuity? 3. It is known that if F n is a prime number, then n must be a prime number. The converse is not true. In fact, the following Fibonacci numbers are composites. Give a factorization of each of them: p F p Factorization

50 14 Fibonacci and Lucas numbers 7.4 The Lucas numbers The sequence (L n ) satisfying L n+ = L n+1 + L n, L 1 =1,L =3, is called the Lucas sequence, and L n the n-th Lucas number. Here are the beginning Lucas numbers. n L n Let α>βbe the roots of the quadratic polynomial x x L n = α n + β n.. L n = F n +F n L n+1 + L n 1 =5F n. 4. F k = L 1 L L 4 L 8 L k L 1 =1and L 3 =4are the only square Lucas numbers (U. Alfred, 1964). Exercise 1. Prove that L n = L n +( 1) n 1. Solution. L n = α n + β n = (α n + β n ) (αβ) n = L n ( 1) n.. Express F 4n F n in terms of L n. Solution. F 4n = F n L n = F n L n L n = F n (L 3 n +( 1) n 1 L n ). 3. Express F 3n F n in terms of L n. Answer. F 3n = F n (L n +( 1) n ).

51 7.5 How can we know if a given integer is a Fibonacci number? How can we know if a given integer is a Fibonacci number? Theorem 7.1 (Gessel). An integer N is a Fibonacci number if and only if 5N ± 4 is a square. Proof. Necessity: 5F n +4( 1) n = L n. Sufficiency: If 5N +4ε = M for ε = ±1 and an integer M, then M + N 5 M N 5 = ε = ±1, and M+N 5 is a unit in Q[ 5]. The units in Q[ 5] are ±α n for n Z. Therefore, M + N 5 and N = F n. = α n = (αn + β n )+(α n β n ) Remark. L n = M, F n 1 = M N and F n+1 = M+N. = L n + F n 5,

52 144 Fibonacci and Lucas numbers 7.6 Periodicity of Fibonacci sequence mod m Let m>1be a given integer. We prove that the sequence (F n (mod m)) is periodic. Let f n = F n (mod m). Since there are m 1 possibilities for the pair (f n,f n+1 ), by the pigeonhole principle, there are h<k m such that (f h,f h+1 )=(f k,f k+1 ). Let r be the least positive integer such that (f h+r,f h+r+1 )=(f h,f h+1 ) for some h. Tracing backward, we have (f r,f r+1 )=(f 0,f 1 )=(0, 1). It follows that the sequence (f n ) is periodic with period r. m Fibonacci numbers modulo m period 0, 1, 1, 0, 1, , 1, 1,, 0,,, 1, 0, 1, , 1, 6 5 0, 1, 0 6 0, 1, 4 7 0, 1, , 1, 1 Exercise Complete the table above and construct an arithmetic progression which does not contain any Fibonacci number.

53 Chapter 8 Counting with Fibonacci numbers 8.1 Squares and dominos In how many ways can a 1 n rectangle be tiled with unit squares and dominos (1 squares)? Suppose there are a n ways of tiling a 1 n rectangle. There are two types of such tilings. (i) The rightmost is tiled by a unit square. There are a n 1 of these tilings. (ii) The rightmost is tiled by a domino. There are a n of these. Therefore, a n = a n 1 + a n. Note that a 1 =1and a =. These are consecutive Fibonacci numbers: a 1 = F and a = F 3. Since the recurrence is the same as the Fibonacci sequence, it follows that a n = F n+1 for every n 1.

54 146 Counting with Fibonacci numbers 8. Fat subsets of [n] A subset A of [n] := {1,...,n} is called fat if for every a A, a A (the number of elements of A). For example, A = {4, 5} is fat but B = {, 4, 5} is not. Note that the empty set is fat. How many fat subsets does [n] have? Solution. Suppose there are b n fat subsets of [n]. Clearly, b 1 =(every subset is fat) and b =3(all subsets except [] itself is fat). Here are the 5 fat subsets of [3]:, {1}, {}, {3}, {, 3}. There are two kinds of fat subsets of [n]. (i) Those fat subsets which do not contain n are actually fat subsets of [n 1], and conversely. There are b n 1 of them. (ii) Let A be a fat subset of m elements and n A. If m =1, then A = {n}. If m>1, then every element of A is greater than 1. The subset A := {j 1:j<n,j A} has m 1 elements, each m 1 since j m for every j A. Note that A does not contain n 1. It is a fat subset of [n ]. There are b n such subsets. We have established the recurrence b n = b n 1 + b n. This is the same recurrence for the Fibonacci numbers. Now, since b 1 = =F 3 and b =3=F 4, it follows that b n = F n+ for every n 1. Exercise 1. (a) How many permutations π :[n] [n] satisfy π(i) i 1, i =1,,...,n? (b) Let π be a permutation satisfying the condition in (a). Suppose for distinct a, b [n], π(a) =b. Prove that π(b) =a.

55 8.3 An arrangement of pennies An arrangement of pennies Consider arrangements of pennies in rows in which the pennies in any row are contiguous, and each penny not in the bottom row touches two pennies in the row below. For example, the first one is allowed but not the second one: How many arrangements are there with n pennies in the bottom row? Here are the arrangements with 4 pennies in the bottom, altogether 13. Solution. Let a n be the number of arrangements with n pennies in the bottom. Clearly a 1 =1, a =, a 3 =5, a 4 =13. A recurrence relation can be constructed by considering the number of pennies in the second bottom row. This may be n 1, n,..., 1, and also possibly none. a n = a n 1 +a n + +(n 1)a 1 +1.

56 148 Counting with Fibonacci numbers Here are some beginning values: n a n =13, =34, =89,. These numbers are the old Fibonacci numbers: a 1 = F 1, a = F 3, a 3 = F 5, a 4 = F 7, a 5 = F 9, a 6 = F 11. From this we make the conjecture a n = F n 1 for n 1. Proof. We prove by mathematical induction a stronger result: a n = F n 1, n a k = F n. k=1 These are clearly true for n =1. Assuming these, we have a n+1 = a n +a n 1 +3a n + + na 1 +1 = (a n + a n 1 + a n + + a 1 ) +(a n 1 +a n + +(n 1)a 1 +1) = F n + a n = F n + F n 1 = F n+1 ; n+1 n a k = a n+1 + k=1 k=1 = F n+1 + F n = F (n+1). a k Therefore, the conjecture is established for all positive integers n. In particular, a n = F n 1.

57 8.4 Counting circular permutations Counting circular permutations Let n 4. The numbers 1,,...,n are arranged in a circle. How many permutations are there so that each number is not moved more than one place? Solution. (a) π(n) =n. There are F n permutations of [n 1] satisfying π(i) i 1. (b) π(n) =1. (i) If π(1) =, then π() = 3,...,π(n 1) = n. (ii) If π(1) = n, then π restricts to a permutation of [,...,n 1] satisfying π(i) i 1. There are F n 1 such permutations. (c) π(n) =n 1. (i) If π(n 1) = n, then π(n ) = n 3,...,π() = 1, π(1) = n. (ii) If π(n 1) = n, then π restricts to a permutation of [1,n ] satisfying π(i) i 1. There are F n 1 such permutations. Therefore, there are altogether F n +(F n 1 +1) = L n +such circular permutations. For n =4, this is L 4 +=

58 Chapter 9 Floor and Ceiling For a real number x, we denote by x the greatest integer not exceeding x, (the floor of x), 1 x the least integer not smaller than x, (the ceiling of x). x the integer nearest to x: x, if x x < 1, x := x, if x x 1. {x}:= x [x] the fractional part of x. 9.1 Clock problems Consider a clock with an hours hand and a minutes hand, graduated by the points 1,,...,1 indicating the hours. At time t (hours), t [0, 1), when the hours hand is at position t, the minutes hand is at position {1t}. Example 9.1. Three easy problems At what times do the hours hand and the minutes hand of a clock (1) align with one another, () point in opposite directions, (3) make right angles with each other? 1 This is also denoted [x] in older books.

59 0 Floor and Ceiling Example 9.. At what times must the hands of a clock be interchanged in order to obtain a new correct time? Solution. Let t [0, 1). At time t (hours), the hours hand is at position t and the minutes hand at position 1{t}. If the two hands are interchanged, the time is s =1{t} hours. The minutes hand should be at position 1{1t}. Now we require t =1{1t}. t = 1(1t [1t]) = 144t 1[1t] = 143t = 1[1t]. Therefore, t = 1k for an integer k =1,..., Some equations involving x Example 9.3. Solve nx = n for a given integer n. If n =0, this is true for all x. If n =1, x =1 = x [1, ). If n = 1, x = 1 = x =1 = x (0, 1]. Now we assume n > 1. Write x = m + t for an integer m and 0 t<1. nx = mn + nt = n = nt =(1 m)n. Since 0 t<1, nt is strictly between n and n. It is a multiple of n only when nt =0and m =1. Therefore, for n>1, 0 t< 1 n, and x = m + t [ 1, 1+ 1 n). For n< 1, t =0and m =1, x =1. Summary: {1}, if n< 1, (0, 1], if n = 1, {x : nx = n} = R, if n =0, [0, 1), if n =1, [ 1, 1+ 1 n), if n>1. Example 9.4. For an integer n, solve the equation n x = nx. Write x = m + t for an integer m and 0 t<1. nm = nm + nt = nm + nt. nt =0 = 0 t< 1 n.

60 9.3 Number of trialing zeros in n! 03 {x : n x = nx } = m Z [ m, m + 1 ). n Example 9.5. Let n be a positive integer. Solve the equation {x } = n {x}. 9.3 Number of trialing zeros in n! (1) Let p be a prime. The highest power of p dividing n! is n n ν p (n!) = + + p p = n s p(n), p 1 where s p (n) is the sum of the digits in the base p expansion of n. () The number of trailing zeros in n! is ν 5 (n) = n s 5(n) 4. Example 9.6. Find the smallest n, if possible, for which n! ends in 015 zeros. Solution. This number is around = Numbers around 8060 has 6 digits in their base 5 expansions. Therefore such a number is between 8060 and = Now, 8074 = has digit sum ! has 1 ( ) = trailing zeros. On the other hand, 8075 = has digit sum ! has 1( ) = 016 trailing zeros. There is no integer n for which n! has exactly 015 trailing zeros. Remark. The following numbers within 100 do not appear as the number of trailing zeros of n! for any integer n: 5, 11, 17, 3, 9, 30, 36, 4, 48, 54, 60, 61, 67, 73, 79, 85, 91, 9, 98,...

61 04 Floor and Ceiling 9.4 Characterizations of the ceiling and floor functions Theorem. Let θ : Q Z be a function satisfying (i) θ(c + x) =c + θ(x) for c Z and x Q, (ii) θ( θ(x) )=θ( x ) for x Q and positive integer n. n n Then either θ(x) = x or θ(x) = x. Proof. (1) If θ(0) = k Z, then k = θ(0) = (ii) θ(θ(0)) = θ(k) =θ(k +0)= (i) k + θ(0). Therefore θ(0) = 0. () For every integer m, θ(m) =θ(m +0)=m + θ(0) = m. (3) Let θ( 1 )=mfor an integer m. For any nonnegative integer k, ( ) ( ( )) ( ( )) 1 1 k +1 θ k + 1 m = θ = θ (ii) k +1 θ = θ k +1 ( k + θ( 1 = θ ) ) ( ) k + m = θ. (ii) k +1 k +1 (i) If m<0, then by putting k = m(> 0), we have m = θ(0) = 0, a contradiction. (ii) If m>1, then by putting k = m 1, we have m = θ(1) = 1, again a contradiction. Therefore, θ ( 1 ) =0or 1. (4) Assume θ ( 1 ) =0. We prove (by induction) that θ ( 1 n) =0for every integer n>1. This is true for n =. Assume θ ( 1 n) =0for some integer n. Then θ ( ) ( ( n+1 n = θ 1+ 1 n) =1+θ 1 n) =1+0=1. It follows that θ ( 1 ) n +1 ( 1 = θ n +1 θ ( )) n +1 = θ n Therefore the proof is complete by induction. ( ) 1 =0. n 1 P. Eisele and K. P. Hadeler, Game of cards, dynamical systems, and a characterization of the floor and ceiling functions, AMERICAN MATH. MONTHLY, 97 (1990)

62 9.4 Characterizations of the ceiling and floor functions 05 (5) We prove (by induction on m) that θ ( ) m n =0for 0 <m<n. By (4) above this is true for m =1. Assume θ ( k n) =0for all k =0, 1,...,m 1. Consider θ ( ) m n. Let n = m(h 1) + r for integers h and r satisfying h 1 and 0 r<m. (i) If r =0, then θ ( ) ( m n = θ 1 h 1) =0by (4). (ii) If r>0, then h 1= n r, h = n+m r, mh = n+m r =1+ m r. m m n n n Therefore, θ ( ) ( ) ( mh n = θ 1+ m r n =1+θ m r ) n =1+0=1. It follows that ( m ) ( ( )) ( ) 1 mh 1 θ = θ n h θ = θ =0. n h This completes the proof of (5) by induction. (6) Summary (assuming θ ( 1 ) =0): if r (0, 1) Q, then θ(r) =0. It follows that for x Q, θ(x) =θ( x + {x}) = x + θ({x}) = x +0= x. (7) It remains to consider the case when θ ( 1 ) =1. If θ satisfies (i) and (ii), then so does θ defined by θ (x) = θ( x). Proof: (i) θ (c + x) = θ( c x) = ( c + θ( x)) = c θ( x) = c + θ (x). ) ( (ii) θ θ (x) n (8) If θ ( 1 ( = θ θ (x) n ) = θ ( ) θ( x) n = θ ( ) ( x n = θ x n). ) =1, then θ ( ) ( ) ( ( ( 1 = θ 1 = θ 1+ 1 ) = 1+θ 1 )) = ( 1+1)= 0. (9) Therefore, θ (x) = x, and θ(x) = θ ( x) = x = x.