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9}, let k be the unique integer such that k ∈ {1, 2, . . , 10} and (j + 1)k ≡ 1 (mod 11). Suppose that [a2m−1 · · · a1 j] ∈ A0 , that is, (a2m−1 , . . , a1 , j) satisfies (∗). From (1) and (2), it follows that (a2m−1 + 1)k − 1, . . , (a1 + 1)k − 1, 0 also satisfies (∗). Putting bi = (ai + 1)k − 1, we have [b2m−1 · · · b1 ] ∈ B0 . For any j ∈ {0, 1, . . , 9}, we can reconstruct [a2m−1 . . a1 j] from [b2m−1 · · · b1 ]. Hence we have |A0 | = 10|B0 |, and so f (2m) = 10f (2m − 1). 31 Geometry G1.

Let s = rt. Then the problem reduces to the following lemma: Lemma. Let b, r, t, t , u be positive integers satisfying b3 − 1 = rtt and t + t = ub2 . Then r = 1. Furthermore, either one of t or t or u is 1. 2 The lemma is proved as follows. We have b3 − 1 = rt(ub√ − t) = rt (ub2 − t ). Since 2 2 2 2 rt ≡ rt ≡ 1 (mod b ), if rt = 1 and rt = 1, then t, t > b/ r. It is easy to see that 2 b b r √ ub2 − √ r r ≥ b3 − 1, unless r = u = 1. 3. With the same notation as in the previous solution, since rt2 | (b3 − 1)2 , it suffices to prove the following lemma: Lemma.

Consider now the case b ≥ 11 and put yn = (b − 1)xn . Assume that the condition in the problem is satisfied. Then it follows that yn yn+1 is a perfect square for n > M . Since b2n + bn+1 + 3b − 5 < (bn + b/2)2 , we infer yn yn+1 < bn + b 2 2 bn+1 + b 2 2 = b2n+1 + bn+1 (b + 1) b2 + 2 4 2 . (1) On the other hand, we can prove by computation that yn yn+1 > b 2n+1 bn+1 (b + 1) − b3 + 2 2 . (2) From (1) and (2), we conclude that for all integers n > M , there is an integer an such that 2 b2 bn+1 (b + 1) 3 2n+1 + an and − b < an < .

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44th International Mathematical Olympiad: Short-listed problems and solutions


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