HMMT 二月 2015 · 冲刺赛 · 第 17 题

HMMT February 2015 — Guts Round — Problem 17

[ 11 ] Find the least positive integer N > 1 satisfying the following two properties: • There exists a positive integer a such that N = a (2 a − 1). • The sum 1 + 2 + · · · + ( N − 1) is divisible by k for every integer 1 ≤ k ≤ 10.

解析

[ 11 ] Find the least positive integer N > 1 satisfying the following two properties: • There exists a positive integer a such that N = a (2 a − 1). • The sum 1 + 2 + · · · + ( N − 1) is divisible by k for every integer 1 ≤ k ≤ 10. 2 Answer: 2016 The second condition implies that 16 divides a (2 a − 1)(2 a − a − 1), which shows that a ≡ 0 or 1 modulo 16. The case a = 1 would contradict the triviality-avoiding condition N > 1. 2 a cannot be 16, because 7 does not divide a (2 a − 1)(2 a − a − 1). a cannot be 17, because 9 does not 2 divide a (2 a − 1)(2 a − a − 1). It can be directly verified that a = 32 is the smallest positive integer 4 2 for which 1 + 2 + · · · + ( N − 1) = 2 · 3 · 5 · 7 · 13 · 31 which is divisible by 1 , 2 , . . . , 10. For this a , we compute N = 32(2 · 32 − 1) = 2016.