HMMT 二月 2003 · 冲刺赛 · 第 11 题

英文解析

11. Find the smallest positive integer n such that 1 + 2 + 3 + 4 + · · · + n is divisibleby 100.
    Solution: 24
    The sum of the first n squares equals n ( n + 1)(2 n + 1) / 6, so we require n ( n + 1)(2 n + 1)
    to be divisible by 600 = 24 · 25. The three factors are pairwise relatively prime, so
    A2
    BCQRone of them must be divisible by 25. The smallest n for which this happens is n = 12DUV
    (2 n + 1 = 25), but then we do not have enough factors of 2. The next smallest isn = 24 ( n + 1 = 25), and this works, so 24 is the answer.