(1901年匈牙利数学竞赛题)
证明: 1^n+2^n+3^n+4^n
能被5整除的充分必要条件是 n 不能被4整除
Let n = 4k+i, where i = 0, 1, 2 or 3 1^(4k+i) mod 5 = 1 2^(4k+i) mod 5 = 16^k * 2^i mod 5 = 2^i 3^(4k+i) mod 5 = 81^k * 3^i mod 5 = 3^i 4^(4k+i) mod 5 = 256^k * 4^i mod 5 = 4^i i = 0, the sum mod 5 = 4 for all other 3 cases, sum mod 5 = 0 |
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