Problem Solving and Recreational Mathematics

Transcription

1 Problem Solving and Recreational Mathematics Paul Yiu Department of Mathematics Florida Atlantic University Summer 2012 Chapters 1 44 August 1 Monday 6/25 7/2 7/9 7/16 7/23 7/30 Wednesday 6/27 *** 7/11 7/18 7/25 8/1 Friday 6/29 7/6 7/13 7/20 7/27 8/3

2 ii

3 Contents 1 Digit problems When can you cancel illegitimately and yet get the correct answer? Repdigits Sorted numbers with sorted squares Sums of squares of digits Transferrable numbers Right-transferrable numbers Left-transferrable integers Arithmetic problems A number game of Lewis Carroll Reconstruction of multiplications and divisions A multiplication problem A division problem Fibonacci numbers The Fibonacci sequence Some relations of Fibonacci numbers Fibonacci numbers and binomial coefficients Counting with Fibonacci numbers Squares and dominos Fat subsets of [n] An arrangement of pennies Fibonacci numbers Factorization of Fibonacci numbers

4 iv CONTENTS 6.2 The Lucas numbers Counting circular permutations Subtraction games The Bachet game The Sprague-Grundy sequence Subtraction of powers of Subtraction of square numbers More difficult games The games of Euclid and Wythoff The game of Euclid Wythoff s game Beatty s Theorem Extrapolation problems What is f(n + 1) if f(k) = 2 k for k = 0, 1, 2..., n? What is f(n + 1) if f(k) = 1 for k = 0, 1, 2..., n?. 315 k Why is e x not a rational function? The Josephus problem and its generalization The Josephus problem Chamberlain s solution The generalized Josephus problem J(n, k) The nim game The nim sum The nim game Prime and perfect numbers Infinitude of prime numbers Euclid s proof Fermat numbers The sieve of Eratosthenes A visualization of the sieve of Eratosthenes The prime numbers below Perfect numbers Mersenne primes Charles Twigg on the first 10 perfect numbers Primes in arithmetic progression

5 CONTENTS v 12.8 The prime number spirals The prime number spiral beginning with The prime number spiral beginning with Cheney s card trick Three basic principles The pigeonhole principle Arithmetic modulo Permutations of three objects Examples Variations of Cheney s card trick Cheney card trick with spectator choosing secret card A 3-card trick The Catalan numbers Number of nonassociative products The golden ratio Division of a segment in the golden ratio The regular pentagon Construction of 36, 54, and 72 angles The most non-isosceles triangle Medians and angle Bisectors Apollonius Theorem Angle bisector theorem The angle bisectors Steiner-Lehmus Theorem Dissections Dissection of the 6 6 square Dissection of a 7 7 square into rectangles Dissect a rectangle to form a square Dissection of a square into three similar parts Pythagorean triangles Primitive Pythagorean triples Rational angles

6 vi CONTENTS Some basic properties of primitive Pythagorean triples A Pythagorean triangle with an inscribed square When are x 2 px ± q both factorable? Dissection of a square into Pythagorean triangles Integer triangles with a 60 or 120 angle Integer triangles with a 60 angle Integer triangles with a 120 angle Triangles with centroid on incircle Construction Integer triangles with centroid on the incircle The area of a triangle Heron s formula for the area of a triangle Heron triangles The perimeter of a Heron triangle is even The area of a Heron triangle is divisible by Heron triangles with sides < Heron triangles with sides in arithmetic progression Indecomposable Heron triangles Heron triangle as lattice triangle Heron triangles Heron triangles with area equal to perimeter Heron triangles with integer inradii Division of a triangle into two subtriangles with equal incircles Inradii in arithmetic progression Heron triangles with integer medians Heron triangles with square areas Triangles with sides and one altitude in A.P Newton s solution The general case The Pell Equation The equation x 2 dy 2 = The equation x 2 dy 2 =

7 CONTENTS vii 25.3 The equation x 2 dy 2 = c Figurate numbers Which triangular numbers are squares? Pentagonal numbers Almost square triangular numbers Excessive square triangular numbers Deficient square triangular numbers Special integer triangles Almost isosceles Pythagorean triangles The generators of the almost isosceles Pythagorean triangles Integer triangles (a, a + 1, b) with a 120 angle Heron triangles Heron triangles with consecutive sides Heron triangles with two consecutive square sides Squares as sums of consecutive squares Sum of squares of natural numbers Sums of consecutive squares: odd number case Sums of consecutive squares: even number case Sums of powers of consecutive integers Lucas problem Solution of n(n + 1)(2n + 1) = 6m 2 for even n The Pell equation x 2 3y 2 = 1 revisited Solution of n(n + 1)(2n + 1) = 6m 2 for odd n Some geometry problems Basic geometric constructions Some basic construction principles Geometric mean Harmonic mean A.M G.M. H.M

8 viii CONTENTS 33 Construction of a triangle from three given points Some examples Wernick s construction problems The classical triangle centers The centroid The circumcircle and the circumcenter The incenter and the incircle The orthocenter and the Euler line The excenters and the excircles The nine-point circle The nine-point circle Feuerbach s theorem Lewis Carroll s unused geometry pillow problem Johnson s theorem Triangles with nine-point center on the circumcircle The excircles A relation among the radii The circumcircle of the excentral triangle The radical circle of the excircles Apollonius circle: the circular hull of the excircles Three mutually orthogonal circles with given centers The Arbelos Archimedes twin circle theorem Incircle of the arbelos Construction of incircle of arbelos Archimedean circles in the arbelos Constructions of the incircle Menelaus and Ceva theorems Menelaus theorem Ceva s theorem Routh and Ceva theorems Barycentric coordinates Cevian and traces Area and barycentric coordinates

9 CONTENTS ix 40 Elliptic curves A problem from Diophantus Dudeney s puzzle of the doctor of physic Group law on y 2 = x 3 + ax 2 + bx + c Applications of elliptic curves to geometry problems Pairs of isosceles triangle and rectangle with equal perimeters and equal areas Triangles with a median, an altitude, and an angle bisector concurrent Integer triangles with an altitude equal to a bisector A quartic equation Transformation of a quartic equation into an elliptic curve 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)

10

11 Chapter 1 Digit problems 1.1 When can you cancel illegitimately and yet get the correct answer? Let ab and bc be 2-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, 26 = 2, 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 = 2ab. The only possibilities are (b, a) = (6, 2), (9, 1), giving distinct (a, b, c) = (1, 9, 5), (2, 6, 5).

12 102 Digit problems 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 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 b ad = c d, (b) c a b d b a = c d, (c) a b c b cd = a d. 2. Find all 4-digit numbers like 1805 = , which, when divided by the its last two digits, gives the square of the number one more than its first two digits.

13 1.2 Repdigits 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). Exercise 1. Show that 16 n 6 n 4 = 1 4, 19 n 9 n 5 = 1 5, 26 n 6 n 5 = 2 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

14 104 Digit problems Exercise 1. Prove that for 1 a, b 9, a b n = b a n. 2. Complete the following multiplication table of repdigits. 1 n 2 n 3 n 4 n 5 n 6 n 7 n 8 n 9 n 1 1 n 2 n 3 n 4 n 5 n 6 n 7 n 8 n 9 n 2 4 n 6 n 8 n 1 n 0 13 n n n n n 13 n n n n n n n n 0 26 n n n n n n 0 38 n n 0 49 n n n n n n n n n n n Verify 26n 6 n5 = 2 5. Solution. It is enough to verify 5 (26 n ) = 2 (6 n 5). 5 (26 n ) = 5(20 n + 6 n ) = 10 n n 0 = 13 n 0; 2 (6 n 5) = 2(6 n 0 + 5) = 13 n = 13 n Simplify (1 n )(10 n 1 5). Answer. (1 n )(10 n 1 5) = 1 n 5 n.

15 1.3 Sorted numbers with sorted squares Sorted numbers with sorted squares A number is sorted if its digits are nondecreasing from left to right. It is strongly sorted if its square is also sorted. It is known that the only strongly sorted integers are given in the table below. 1 1, 2, 3, 6, 12, 13, 15, 16, 38, 116, n 7. 3 n 4. 3 n 5. 3 m 6 n 7. (3 n 5 1 ) 2 =(10 3 n + 5) 2 =100 (3 n ) (3 n ) + 25 =1 n 1 08 n n 25 =1 n 1 12 n =1 n 2 n+1 5. If x = 3 m 6 n 7, then 3x = 10 m 1 10 n 1, and it is easy to find its square. (3 m 6 n 7) 2 = { 1 m 3 m 4 n m+1 6 m 8 n 9, if n + 1 m, 1 m 3 n+1 5 m n 1 6 n+1 8 n 9, if n + 1 < m. More generally, the product of any two numbers of the form 3 m 6 n 7 is sorted. 1 Problem 1234, Math. Mag., 59 (1986) 1, solution, 60 (1987)1. See also R. Blecksmith and C. Nicol, Monotonic numbers, Math. Mag., 66 (1993)

16 106 Digit problems Exercise 1. Find all natural numbers whose square (in base 10) is represented by odd digits only. 2. 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 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, 2, 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? 5. 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? 6. 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, 259, 296. 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 < 27. We need only consider k = 4, 5, 7, 8.

17 1.3 Sorted numbers with sorted squares 107 k 37 k s 37 s arithmetic progression = , 481, = , 518, = , 592, = , 629, 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? 8. Find the smallest natural number N, such that, in the decimal notation, N and 2N together use all the ten digits 0, 1,..., 9. Answer. N = and 2N =

18 108 Digit problems 1.4 Sums of squares of digits Given a number N = a 1 a 2 a n of n decimal digits, consider the sum of digits function For example s(n) = a a a2 n. s(11) = 2, s(56) = 41, s(85) = 89, s(99) = 162. For a positive integer N, consider the sequence S(N) : N, s(n), s 2 (N),...,s k (N),..., where s k (N) is obtained from N by k applications of s. Theorem 1.1. For every positive integer N, the sequence S(N) is either eventually constant at or periodic. The period has length 8 and form a cycle Exercise 1. Prove by mathematical induction that 10 n 1 > 81n for n 4. Solution. Clearly this is true for n = 4: 10 3 > Assume 10 n 1 > 81n. Then 10 n = n 1 > 10 81n > 81(n + 1). Therefore, 10 n 1 > 81n for n 4.

19 1.4 Sums of squares of digits 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. N. We seek all 3-digit numbers N = abc for which s(n) (i) Since s(n) 243, we need only consider n 243. (ii) Now if a = 2, then s(n) = 4 + b 2 + c < N. Therefore a = 1. (iii) s(n) = 1 2 +b 2 +c 2 is a 3-digit number if and only if b 2 +c 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 2-digit number. 5. If n is one of the integers 1, 7, 10, 13, 19, 23, 28, 31, 32,44, 49, 68, 70,79, 82, 86, 91,94, 97, then s k (N) = 1 for some k

20 110 Digit problems 6. If N is a 2-digit integer other than 1, 7, 10, 13, 19, 23, 28, 31, 32, 44, 49,68,70, 79, 82,86,91, 94, 97, then the sequence S(N) is eventually cycling between 4, 16, 37, 58, 89, 145, 42,

21 Chapter 2 Transferrable numbers 2.1 Right-transferrable numbers A positive integer is right-transferrable if in moving its leftmost digit to the rightmost position results in a multiple of the number. Suppose a right-transferrable number X has n digits, with leftmost digit a. We have 10(X a 10 n 1 ) + a = kx for some integer k satisfying 1 k 9. From this, (10 k)x = a(10 n 1) = 9 a n, and X = 9 a 1 n 10 k. Clearly, k = 1 if and only if X = a n, a repdigit. We shall henceforth assume k > 1. Since X is an n-digit number, we must have a < 10 k. Most of the combinations of (a, k) are quickly eliminated. In the table below, N indicates that X is not an integer, and R indicates that X is a repdigit (so that k cannot be greater than 1).

22 112 Transferrable numbers k \ a N N N N N N N 3 4 N R N R N 5 N N N N 6 N N N 7 R R 8 N 9 This table shows that k must be equal to 3, and X = a(10n 1). 7 Since a < 7, we must have 7 dividing 10 n 1. This is possible only if n is a multiple of 6. Therefore X = a 106m 1 and has first digit a. 7 Now, 10 6m 1 = (142857) m. 7 It is easy to see that a can only be 1 or 2. Therefore, the only right-transferrable numbers are (142857) m and (285714) m with k = 3: 3 (142857) m = (428571) m, 3 (285714) m = (857142) m.

23 2.2 Left-transferrable integers Left-transferrable integers A positive integer is left-transferrable if moving its rightmost digit to the leftmost results in a multiple of the number. Suppose Y is a lefttransferrable with rightmost digit b. Then b 10 n 1 + Y b 10 = ky for an integer k. From this, (10k 1)Y = b(10 n 1). Again, k = 1 if and only if Y = b n. We shall assume k > 1. Note that for k = 2, 3, 6, 8, 9, p = 10k 1 is a prime number p > 10 > b. It divides 10 n 1. By Fermat s theorem, the order of 10 mod p is a divisor of p 1. k n 2 19Y = b(10 n 1) n 1 18m 3 29Y = b(10 n 1) n 1 28m 4 39Y = b(10 n 1) b 3, 6, 9; n 1 6m 5 49Y = b(10 n 1) b 7; n 1 42m b = 7; 7 10 n 1 6m 6 59Y = b(10 n 1) n 1 58m 7 69Y = b(10 n 1) b 3, 6, 9; n 1 22m 8 79Y = b(10 n 1) n 1 13m 9 89Y = b(10 n 1) n 1 44m Consider the case k = 2. We have n = 18m. If we take n = 18, then Y = b(1018 1) 19 = b Note that A = = has only 17 nonzero digits, we treat it as an 18-digit number For each b = 1,...,9, the number Y b = b(1018 1) 19 is right-transferrable with k = 2:

24 114 Transferrable numbers b b A rightmost digit to leftmost More generally, for each of these Y b and arbitrary positive integer m, Y b,m = Y b ((10 17 ) m 1 )1 = (Y b ) m is also left-transferrable with k = 2. Proof. Write Y b = X b b for a 17-digit number X b. Transferring the rightmost digit of (Y b ) m to the leftmost, we have (bx b ) m = (2 Y b ) m = 2 (Y b ) m. The same holds for the other values of k > 1 as shown in the table below, except for k = 5, where we also have = k p

25 2.2 Left-transferrable integers 115 Exercise 1. What digits should be substituted for the letters so that the sum of the nine identical addends will be a repunit? R E P U N I T S R E P U N I T S R E P U N I T S R E P U N I T S R E P U N I T S R E P U N I T S R E P U N I T S R E P U N I T S + R E P U N I T S 2. Are two repunits with consecutive even numbers as their subscripts relatively prime? 3. Are two repunits with consecutive numbers as their subscripts relatively prime? 4. Are two repunits with consecutive odd numbers as their subscripts relatively prime? 5. What digit does each letter of this multiplication represent? R R R R R R R R R R R R R R R E P U N I T I N U P E R 6. An old car dealer s record in the 1960 s shows that the total receipts for the sale of new cars in one year came to 1, 111, dollars. If each car had eight cylinders and was sold for the same price as each other car, how many cars did he sell? (Note: This riddle was written before the inflation in the 1980 s). 7. If a Mersenne number M p = 2 p 1 is prime, is the corresponding repunit 1 p also prime?

26 116 Transferrable numbers

27 Chapter 3 Arithmetic problems 3.1 A number game of Lewis Carroll How would you get A from D? Take a secret number Multiply it by 3 Tell me if it is Do the corresponding routine as instructed below. Multiply by 3 Tell me if it is Do the corresponding routine Add 19 to the original number A and put an extra digit at the end Now add B and C Divide by 7 and get the quotient only Further divide by 7 and get the quotient only Tell me this and I shall give you back A even or odd even or odd B C D D your A odd routine: Add 5 or 9, then divide by 2, and then add 1. even routine: Subtract 2 or 6, then divide by 2, and then add 29 or 33 or 37.

28 118 Arithmetic problems Solution. A can be obtained from D by (i) forming 4D 15, (ii) subtracting 3 if the first parity answer is even, and (iii) subtracting 2 if the second parity answer is even. Analysis. e and e are either 0 or 1. f and f are 1, 0, or 1. g is an integer between 0 and 9. The last two steps of dividing by 7 and keeping the quotients can be combined into one single step of dividing by 49. A 4k + 1 4k + 2 4k + 3 4k + 4 3A 12k k k k + 12 Parity odd even odd even Routine 6k e 6k k e 6k e + 4f 2e + 4f 3 times 18k e 18k k e e + 12f 6e + 4f Parity odd odd even even Routine B 9k k k k e + 2e 3e + 2e +3e 2e 3e 2e +6f +4f +6f + 4f C 40k g 40k g 40k g 40k g B + C 49k g 49k g 49k g 49k g +3e + 2e 3e + 2e +3e 2e 3e 2e +6f +4f +6f + 4f lower bound 49k k k k upper bound 49k k k k D k + 4 k + 5 k + 5 k + 6 A 4D 15 4D 18 4D 17 4D 20

29 3.1 A number game of Lewis Carroll 119 Exercise 1. There is a list of n statements. For k = 1, 2,..., n, the k-th statement reads: The number of false statements in this list is greater than k. Determine the truth value of each of the statements. Answer. n must be an odd number. Write n = 2m + 1. Statements 1,..., m are true, and statements m+1,..., n are false. 2. A man rowing upstream passes a log after a miles, then continues for b hours, and then rows downstream, meeting the log at his starting point. What is the rate of the stream? 3. If a man takes h hours to make a certain trip, how much faster must he travel to make a trip m miles longer in the same time? 4. The ratio of the speeds of two trains is equal to the ratio of the time they take to pass each other going in the same direction to the time they take to pass each other in the opposite direction. Find the ratio of the speeds of the two trains. 5. You and I are walking toward each other along a straight road, each at a steady speed. A truck (also traveling at a steady speed) passes you in one second, and one second later it reaches me. One second after the truck has passed me, you and I meet. How long does the truck take to pass me? 6. The two hands of a clock have the same length. One can, nevertheless, normally tell the correct time. For example, when the hands point at 6 and 12, it must be 6 O clock, and cannot be otherwise. In every 12 hours period, there are, however, a number of occasions when it is impossible to tell the time. Exactly how many such occasions are there?

30 120 Arithmetic problems 3.2 Reconstruction of multiplications and divisions A multiplication problem A multiplication of a three-digit number by 2-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

31 3.2 Reconstruction of multiplications and divisions A division problem 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 2-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 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 992 = )

32 122 Arithmetic problems Exercise Reconstruct the following division problems. 1. ) 2 2. ) 9 3. 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

33 3.2 Reconstruction of multiplications and divisions 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

34 Chapter 4 Fibonacci numbers 4.1 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 2 x 2 = F 1 x x + F 0 x 2, F 3 x 3 = F 2 x 2 x + F 1 x x 2, F 4 x 4 = F 3 x 3 x + F 2 x 2 x 2,. F n x n = F n 1 x n 1 x + F n 2 x n 2 x 2,.

35 202 Fibonacci numbers Combining these relations we have F(x) (F 0 + F 1 x) = (F(x) F 0 ) x + F(x) x 2, F(x) x = F(x) x + F(x) x 2, (1 x x 2 )F(x) = x. Thus, we obtain the generating function of the Fibonacci numbers: x F(x) = 1 x x 2. There is a factorization of 1 x x 2 by making use of the roots of the quadratic polynomial. Let α > β be the two roots. We have More explicitly, α = α + β = 1, αβ = , β =. 2 2 Now, since 1 x x 2 = (1 αx)(1 βx), we have a partial fraction decomposition x 1 x x = x 2 (1 αx)(1 βx) = 1 ( 1 α β 1 αx 1 ). 1 βx 1 Each of making use of and 1 1 αx 1 βx has a simple power series expansion. In fact, 1 1 x = 1 + x + x2 + + x n + = and noting that α β = 5, we have ( x 1 x x = 1 α n x n 2 5 n=0 α n β n = x n. 5 n=0 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, 2,....

36 4.1 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 2, F n+1 = {αf n }. Proof. Note that For n 2, F n+1 αf n = αβn β n+1 = F n+1 αf n = β n < 1 2. α β 5 β n = β n. 3. 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 is a Fibonacci number. 2. Prove by mathematical induction the Cassini formula: F n+1 F n 1 F 2 n = ( 1)n. 3. The conversion from miles into kilometers can be neatly expressed by the Fibonacci numbers. miles kilometers

37 204 Fibonacci numbers 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? 4. Prove the Fermat Last Theorem for Fibonacci numbers: there is no solution of x n + y n = z n, n 2, in which x, y, z are (nonzero) Fibonacci numbers. 4.2 Some relations of Fibonacci numbers 1. Sum of consecutive Fibonacci numbers: n F k = F n+2 1. k=1 2. Sum of consecutive odd Fibonacci numbers: n F 2k 1 = F 2n. k=1 3. Sum of consecutive even Fibonacci numbers: n F 2k = F 2n+1 1. k=1 4. Sum of squares of consecutive Fibonacci numbers: n Fk 2 = F nf n+1. k=1 5. Cassini s formula: F n+1 F n 1 F 2 n = ( 1)n.

38 4.3 Fibonacci numbers and binomial coefficients Fibonacci numbers and binomial coefficients Rewriting the generating function x 1 x x 2 = x 1 (x+x 2 ) as x + x(x + x 2 ) + x(x + x 2 ) 2 + x(x + x 2 ) 3 + x(x+ 2 ) 4 + = x + x 2 (1 + x) + x 3 (1 + x) 2 + x 4 (1 + x) 3 + x 5 (1 + x) 4 + = x + x 2 + x 3 + x 4 + x x 3 + 2x 4 + 3x 5 + 4x x 5 + 3x 6 + 6x x 7 + 4x x we obtain the following expressions of the Fibonacci numbers in terms of the binomial coefficents: F 1 = F 2 = F 3 = F 4 = F 5 = F 6 = F 7 = F 8 =. ( ) 0 0 = 1, ( ) 1 0 = 1, ( ) ( ) = 2, ( ) ( ) = = 3, ( ) ( ) ( ) = = 5, ( ) ( ) ( ) = = 8, ( ) ( ) ( ) ( ) = = 13, ( ) ( ) ( ) ( ) = = 21,

39 206 Fibonacci numbers Theorem 4.1. For k 0, F k+1 = k 2 ( ) k j. j j=0 Proof. x 1 x x = x (x + x 2 ) n 2 = = = = n=0 x n+1 (1 + x) n n=0 n ( ) n x n+1 x m m m=0 n ( ) n x n+m+1 m m=0 k 2 ( ) k j x k+1. j n=0 n=0 k=1 j=0

40 Chapter 5 Counting with Fibonacci numbers 5.1 Squares and dominos In how many ways can a 1 n rectangle be tiled with unit squares and dominos (1 2 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 2 of these. Therefore, a n = a n 1 + a n 2. Note that a 1 = 1 and a 2 = 2. These are consecutive Fibonacci numbers: a 1 = F 2 and a 2 = F 3. Since the recurrence is the same as the Fibonacci sequence, it follows that a n = F n+1 for every n 1.

41 208 Counting with Fibonacci numbers 5.2 Fat subsets of [n] A subset A of [n] := {1, 2..., 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 = {2, 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 = 2 (every subset is fat) and b 2 = 3 (all subsets except [2] itself is fat). Here are the 5 fat subsets of [3]:, {1}, {2}, {3}, {2, 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 2]. There are b n 2 such subsets. We have established the recurrence b n = b n 1 + b n 2. This is the same recurrence for the Fibonacci numbers. Now, since b 1 = 2 = F 3 and b 2 = 3 = F 4, it follows that b n = F n+2 for every n 1. Exercise 1. (a) How many permutations π : [n] [n] satisfy π(i) i 1, i = 1, 2,..., 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.

42 5.3 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 2 = 3, 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 2,..., 1, and also possibly none. a n = a n 1 + 2a n (n 1)a

43 210 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 2 = 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 2n 1 for n 1. Proof. We prove by mathematical induction a stronger result: a n = F 2n 1, n a k = F 2n. k=1 These are clearly true for n = 1. Assuming these, we have a n+1 = a n + 2a n 1 + 3a n na = (a n + a n 1 + a n a 1 ) + (a n 1 + 2a n (n 1)a 1 + 1) = F 2n + a n = F 2n + F 2n 1 = F 2n+1 ; n+1 n a k = a n+1 + k=1 k=1 = F 2n+1 + F 2n = F 2(n+1). a k Therefore, the conjecture is established for all positive integers n. In particular, a n = F 2n 1.

44 Chapter 6 Fibonacci numbers Factorization of Fibonacci numbers 1. gcd(f m, F n ) = F gcd(m,n). 2. If m n, then F m F n. 3. Here are the beginning values of F 2n F n : n F n F 2n L n = F 2n F n (a) These quotients seem to satisfy the same recurrence as the Fibonacci numbers: each number is the sum of the preceding two. (b) Therefore, it is reasonable to expect that each of these quotients can be expressed in terms of Fibonacci numbers.

45 212 Fibonacci numbers 3 n F n F 2n L n = F 2n F n F n 1 + F n = = = = = = = (c) Conjectures: (a) L n+2 = L n+1 + L n, L 1 = 1, L 2 = 3; (b) L n = F n+1 + F n These conjectures are true. They are easy consequences of Theorem 6.1 (Lucas Theorem). F 2n = F 2 n+1 F 2 n 1, F 2n+1 = F 2 n+1 + F 2 n. Proof. We prove this by mathematical induction. These are true for n = 1. Assume these hold. Then F 2n+2 = F 2n+1 + F 2n = (Fn Fn) 2 + (Fn+1 2 Fn 1) 2 = (Fn Fn+1) 2 + (Fn 2 Fn 1) 2 = Fn Fn (F n + F n 1)(F n F n 1) = Fn Fn F n+1(f n F n 1) = Fn F n+1(f n+1 + (F n F n 1)) = Fn F n+1(f n + (F n+1 F n 1)) = Fn F n+1f n = Fn F n+1f n + Fn 2 Fn 2 = Fn+2 2 Fn; 2 F 2n+3 = F 2n+2 + F 2n+1 = (Fn+2 2 Fn) 2 + (Fn Fn) 2 = Fn Fn+1. 2

46 6.1 Factorization of Fibonacci numbers 213 Therefore the statements are true for all n. 5. By Lucas theorem, F 2n = F 2 n+1 F 2 n 1 = (F n+1 F n 1 )(F n+1 +F n 1 ) = F n (F n+1 +F n 1 ). If we put L n = F n+1 + F n 1, then L 1 = F 2 + F 0 = 1, L 2 = F 1 + F 3 = = 3, and L n+2 = F n+3 +F n+1 = (F n+2 +F n )+(F n+1 +F n 1 ) = L n+1 +L n. 6. If F n is prime, then n is prime. The converse is not true. Of course, F 2 = 1 is not a prime. What is the least odd prime p for which F p is not prime? 7. Apart from F 0 = 0 and F 1 = F 2 = 1, there is only one more Fibonacci number which is a square. What is this?

47 214 Fibonacci numbers The Lucas numbers The sequence (L n ) satisfyin L n+2 = L n+1 + L n, L 1 = 1, L 2 = 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 > β be the roots of the quadratic polynomial x 2 x L n = α n + β n. 2. L n+1 + L n 1 = 5F n. 3. F 2 k = L 1 L 2 L 4 L 8 L 2 k L 1 = 1 and L 3 = 4 are the only square Lucas numbers (U. Alfred, 1964). Exercise 1. Prove that L 2n = L 2 n + 2( 1)n 1. Solution. L 2n = α 2n + β 2n = (α n + β n ) 2 2(αβ) n = L 2 n 2( 1) n. 2. Express F 4n F n in terms of L n. Solution. F 4n = F 2n L 2n = F n L n L 2n = F n (L 3 n + 2( 1)n 1 L n ). 3. Express F 3n F n in terms of L n. Answer. F 3n = F n (L 2 n + ( 1) n ).

48 6.3 Counting circular permutations Counting circular permutations Let n 4. The numbers 1, 2,..., 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) = 2, then π(2) = 3,..., π(n 1) = n. (ii) If π(1) = n, then π restricts to a permutation of [2,..., 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 2, then π(n 2) = n 3,..., π(2) = 1, π(1) = n. (ii) If π(n 1) = n, then π restricts to a permutation of [1, n 2] satisfying π(i) i 1. There are F n 1 such permutations. Therefore, there are altogether F n + 2(F n 1 + 1) = L n + 2 such circular permutations. For n = 4, this is L =

49 Chapter 7 Subtraction games 7.1 The Bachet game Beginning with a positive integer, two players alternately subtract a positive integer < d. The player who gets down to 0 is the winner. There is a set of winning positions in the form of a decreasing sequence of nonnegative integers, such that if you secure one of these positions, then your opponent cannot secure any of the winning positions, and no matter how he moves, you can always secure a (smaller) winning positions. By keeping track of these winning positions, you eventually secure 0 and win the game. In the present example (the Bachet game), the winning positions are precisely the multiples of d. Proof. If Player A occupies position kd, and his opponent subtract a < d, then A subtracts (d a) and occupies occupies position (k 1)d. The same strategy allows A to get to 0 through the multiples of d.

50 302 Subtraction games 7.2 The Sprague-Grundy sequence Let G be a two-person counter game in which two players alternately remove a positive amount of counters according to certain specified rules. The Sprague-Grundy sequence of G is the sequence (g(n)) of nonnegative integers defined recursively as follows. (1) g(n) = 0 for all n which have no legal move to another number. In particular, g(0) = 0. (2) Suppose from position n it is possible to move to any of positions m 1, m 2,..., m k, (all < n), then g(n) is the smallest nonnegative integer different from g(m 1 ), g(m 2 ),..., g(m k ). Theorem 7.1. The player who secures a position n with g(n) = 0 has a winning strategy. Example 7.1. The Bachet game: n g(n) , , 1, 0 3. d 1 (d 2),...,1, 0 d 1 d (d 1),...,1 0. More generally, g(kd + a) = a for integers k and a satisfying 0 a < d. The Sprague-Grundy sequence is the periodic sequence with period 0, 1,..., d 1. The winning positions are precisely the multiples of d. Example 7.2. The trivial counter game. Two players alternately subtract any positive amount. The only winning position is 0. The first player wins by removing all counters. In this case, g(n) = n for every n.

51 7.3 Subtraction of powers of Subtraction of powers of 2 n g(n) , , , 2, , 3, , 4, , 5, , 6, 4, , 7, 5, , 8, 6, 2 1 This suggests that the winning positions are the multiples of 3. Proof. If Player A occupies a multiple of 3, any move by Player B will results in a position 3k + 1 or 3k + 2. Player A can get to a smaller multiple of 3 by subtracting 1 or 2 accordingly. Exercise 1. What are the winning positions in the game of subtraction of powers of 3? Answer. Even numbers. 2. What are the winning positions in the game of subtraction of prime numbers or 1? Answer. Multiples of What are the winning positions in the game of subtraction of a proper divisor of the current number (allowing 1 but not the number itself)? Note that 1 is not a proper divisor of itself Answer. Odd numbers except 1. Solution. The clue is that all factors of an odd number are odd. Subtracting an odd leaves an even number. Hence the winning strategy is to leave an odd number so that you opponent will always leave you an even number. From this you get to an odd number by subtracting 1.

52 304 Subtraction games 7.4 Subtraction of square numbers Two players alternately subtract a positive square number. We calculate the Sprague-Grundy sequence. n g(n) , , , , , , 5, , 6, 1 0 The values of n 500 for which g(n) = 0 are as follows: 1 These are the winning positions Suppose we start with 74. Player A can subtract 9 to get to 65, or subtract 64 to get 10, which have value 0. In the latter case, B may move to 9, 6 or 1. A clearly wins if B moves to 9 or 1. But if B moves to 6, then A can move to 5 or 2 and win. Exercise 1. How would you win if the starting number is 200? or 500? 1 [Smith, p.68] incorrectly asserts that this sequence is periodic, with period 5.

53 7.5 More difficult games More difficult games 1. Subtraction of proper divisor of current number (not allowing 1 and the number itself). The winning positions within 500 are as follows Subtraction of primes (not allowing 1). The winning positions within 500 are as follows

54 306 Subtraction games

55 Chapter 8 The games of Euclid and Wythoff 8.1 The game of Euclid Two players alternately remove chips from two piles of a and b chips respectively. A move consists of removing a multiple of one pile from the other pile. The winner is the one who takes the last chip in one of the piles. Preliminary problem: Find a constant k such that for positive integers a and b satisfying b < a < 2b, (i) a < kb = b > k(a b), (ii) a > kb = b < k(a b). Analysis: If these conditions hold, then we have a = kb = b = k(a b). If a = k, then 1 = k ( a 1) = k(k 1). Such a number k b b must satisfy k(k 1) = 1; it is the golden ratio ϕ = 1 ( 5 + 1). 2 Proposition 8.1. Let ϕ := 1 2 ( 5 + 1). For any two real numbers a and b, (i) a < ϕb = b > ϕ(a b), (ii) a > ϕb = b < ϕ(a b). Proof. The golden ratio ϕ satisfies ϕ 1 = 1. ϕ (i) a < ϕb = a b < (ϕ 1)b = 1 b = ϕ(a b) < b. ϕ (ii) a > ϕb = a b > (ϕ 1)b = 1 b = ϕ(a b) > b. ϕ Theorem 8.2. In the game of Euclid (a, b), the first player has a winning strategy if and only if a > ϕb.

56 308 The games of Euclid and Wythoff Proof. The first player (A) clearly wins if a = kb for some integer k. Assume a > ϕb. Let q be the largest integer such that a > qb. If q = 1, the only move is (a, b) (b, a b). In this case, b < ϕ(a b) by Proposition 8.1(ii). If q 2, we consider the moves (i) (a, b) (b, a qb) and (ii) (a, b) (a (q 1)b, b) (Note that a (q 1)b > b). If b < ϕ(a qb), make move (i). Otherwise, b > ϕ(a qb). Make move (ii). In this case, a qb < 1 ϕ b = a (q 1)b < ( 1 ϕ + 1 ) b = ϕb. This means that A can make a move (a, b) (a, b ) with a > b such that a < ϕb. The second player B has no choice but only the move (a, b ) (a b, b ). By Proposition 8.1, b > ϕ(a b ). Assume a < ϕb. Then A has no choice except the move (a, b) (a b, b). Here, b > ϕ(a b). Winning strategy: Suppose a > ϕb. Let q be the largest integer such that a > qb. A chooses between (a, b ) = (b, a qb) or (a (q 1)b, b) for a < ϕb. Examples: (a,b) A moves to B moves to (50, 29) (29, 21) (21, 8) (8,5) (5,3) (3,2) (2,1) (1,0) wins (a,b) A moves to B moves to (50, 31) (31, 19) (19, 12) (12, 7) (7, 5) (5,3) (3,2) (2,1) (1,0) wins

57 8.2 Wythoff s game Wythoff s game Given two piles of chips, a player either removes an arbitrary positive amount of chips from any one pile, or an equal (positive) amount of chips from both piles. The player who makes the last move wins. We describe the position of the game by the amounts of chips in the two piles. If you can make (1, 2), then you will surely win no matter how your opponent moves. Now, to forbid your opponent to get to this position, you should occupy (3, 5). The sequence of winning positions: starting with (a 1, b 1 ) = (1, 2), construct (a n, b n ) by setting a n := min{c : c a i, b i, i < n}, b n :=a n + n. Here are the 18 smallest winning positions for Wythoff s game: Every positive integer appears in the two sequences. Proof. Suppose (for a contradiction) that not every positive integer appears in the two sequences. Let N be the smallest of such integers. There is a sufficiently large integer M such that all integers less than N are among a 1,...,a M and b 1,...,b M. Then by definition, a M+1 = N, a contradiction. 2. The sequence (a n ) is increasing. Proof. Suppose a n+1 a n. This means that a n+1 is the smallest integer not in the list a 1, a 2,...,a n 1, a n, b 1, b 2,...,b n 1, b n. In particular, a n+1 a n. This means that a n+1 < a n < a n + n = b n. Ignoring a n and b n from this list, a n+1 is the smallest integer not in a 1, a 2,...,a n 1, b 1, b 2,...,b n 1. This means, by definition, that a n = a n+1, a contradiction.