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Number theory Difficulty 4.7 AIME Prove it Slovenia

Let mm and nn be positive integers such that 5m+n5m + n divides 5n+m5n + m. Prove that mm divides nn.

Solution

There exists a positive integer kk such that 5n+m=k(5m+n)5n + m = k(5m + n), or (5k)n=(5k1)m(5 - k)n = (5k - 1)m. The right-hand side is strictly positive, so the left-hand side must be strictly positive as well, which implies 5k>05 - k > 0 and so k{1,2,3,4}k \in \{1, 2, 3, 4\}. If k=1k = 1, then 4n=4m4n = 4m, so n=mn = m. If k=2k = 2, then 3n=9m3n = 9m, so n=3mn = 3m. If k=3k = 3, then 2n=14m2n = 14m, so n=7mn = 7m. If k=4k = 4, then n=19mn = 19m. In each case mm divides nn.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.