HMMT 十一月 2018 · 团队赛 · 第 5 题

HMMT November 2018 — Team Round — Problem 5

[ 35 ] Find the sum of all positive integers n such that 1 + 2 + · · · + n divides 2 2 2 15[( n + 1) + ( n + 2) + · · · + (2 n ) ] .

解析

[ 35 ] Find the sum of all positive integers n such that 1 + 2 + · · · + n divides 2 2 2 15[( n + 1) + ( n + 2) + · · · + (2 n ) ] . Proposed by: Michael Ren Answer: 64 n ( n +1) 2 n (2 n +1)(4 n +1) 2 2 2 We can compute that 1 + 2 + · · · + n = and ( n + 1) + ( n + 2) + · · · + (2 n ) = − 2 6 n ( n +1)(2 n +1) n (2 n +1)(7 n +1) 15(2 n +1)(7 n +1) 5(2 n +1)(7 n +1) = , so we need = to be an integer. The remain- 6 6 3( n +1) n +1 30 der when (2 n + 1)(7 n + 1) is divided by ( n + 1) is 6, so after long division we need to be an integer. n +1 The solutions are one less than a divisor of 30 so the answer is 1 + 2 + 4 + 5 + 9 + 14 + 29 = 64 .