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**Extra info for Mathematical Olympiads, Problems and Solutions from Around the World, 1996-1997**

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Solution: We dissect a 7 × 7 square into a 2 × 2 square A, a 3 × 3 square B, and three pieces C, D, E which form a 6 × 6 square, as shown below. C C C C C A A C C C C C A A C C C C C D D C C C C C D D C C C C B B B C C C C B B B E E E E B B B 6. Let Fn denote the Fibonacci sequence, so that F0 = F1 = 1 and Fn+2 = Fn+1 + Fn for n ≥ 0. Prove that (i) The statement “Fn+k − Fn is divisible by 10 for all positive integers n” is true if k = 60 and false for any positive integer k < 60; (ii) The statement “Fn+t − Fn is divisible by 100 for all positive integers n” is true if t = 300 and false for any positive integer t < 300.

Hence the eight points maintain their relative position under f , which suffices to prove isometry. 14. Let n ≥ 3 be an integer and X ⊆ {1, 2, . . , n3 } a set of 3n2 elements. Prove that one can find nine distinct numbers a1 , . . , a9 in X such that the system a1 x + a2 y + a3 z a4 x + a5 y + a6 z a7 x + a8 y + a9 z = 0 = 0 = 0 has a solution (x0 , y0 , z0 ) in nonzero integers. Solution: Label the elements of X in increasing order x1 < · · · < x3n2 , and put X1 = {x1 , . . , xn2 }, X2 = {xn2 +1 , .

The period mod 25 turns out to be 100, which is awfully many terms to compute by hand, but knowing that the period must be a multiple of 20 helps, and verifying the recurrence Fn+8 = tFn+4 + Fn , where t is an integer congruent to 2 modulo 5, shows that the period divides 100; finally, an explicit computation shows that the period is not 20. 7. Prove that for all positive integers n, n 21/2 · 41/4 · · · (2n )1/2 < 4. Solution: It suffices to show ∞ ∞ n = 2n n=1 n=1 ∞ k=n ∞ n=1 n/2n = 2: ∞ 1 1 = = 2.