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Solution: If s is a power of 2, then there is nothing to prove. If s is not a power of 2 then it must lie between two consecutive powers of 2, say 2r < s < 2r+1 . This yields 2r+1 < 2s. Hence s < 2r+1 < 2s, which yields the result. 250 Definition The Fibonacci Numbers are given by f0 = 0, f1 = 1, fn+1 = fn + fn−1 , n ≥ 1, that is every number after the second one is the sum of the preceding two. The Fibonacci sequence then goes like 0, 1, 1, 2, 3, 5, 8, 13, 21, . .. 251 Example Prove that for integer n ≥ 1, fn−1 fn+1 = fn2 + (−1)n+1 .

9 55 00 . . 0 . ßÞ ßÞ n−3 9 s n−2 0 s 215 Problem Shew that for any positive integer n, 11 . . 1 − 22 . . 2 ßÞ ßÞ 2n 1 s n2s acb, bac, cab, cba, to add these numbers and to reveal their sum N. If told the value of N, the magician can identify abc. Play the magician and determine abc if N = 319. 224 Problem (AIME 1988) For any positive integer k, let f1 (k) denote the square of the sums of the digits of k. For n ≥ 2, let fn (k) = f1 ( fn−1 (k)). Find f1988 (11). 225 Problem (IMO 1969) Determine all three-digit numbers N that are divisible by 11 N equals the sum of the squares of the digits of N.

A= a6 + 45a5 − 81a2 + 27 6a(a2 + 3)2 3 −a2 + 30a2 − 9 + 6a(a2 + 3) 3 −6a3 + 18a + . (a2 + 3)2 3 331 Problem Let k ≥ 2 be an integer. Shew that if n is a positive integer, then nk can be represented as the sum of n successive odd numbers. 332 Problem (IMO 1979) If a, b are natural numbers such that 1 1 1 1 1 a = 1 − + − + ··· − + , b 2 3 4 1318 1319 If a is rational this shews that every rational number is expressible as the sum of the cubes of three rational numbers. prove that 1979|a. 320 Problem What is the largest power of 7 that divides 1000!?