HMMT 二月 2007 · 冲刺赛 · 第 27 题

HMMT February 2007 — Guts Round — Problem 27

[ 12 ] Find the number of 7-tuples ( n , . . . , n ) of integers such that 1 7 7 ∑ 6 n = 96957 . i i =1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

解析

[ 12 ] Find the number of 7-tuples ( n , . . . , n ) of integers such that 1 7 7 ∑ 6 n = 96957 . i i =1 Answer: 2688 . Consider the equation in modulo 9. All perfect 6th powers are either 0 or 1. Since 9 divides 96957, it must be that each n is a multiple of 3. Writing 3 a = n and dividing both sides by i i i 6 6 6 3 , we have a + · · · + a = 133 . Since sixth powers are nonnegative, | a | ≤ 2. Again considering modulo i 1 7 6 9, we see that a 6 = 0. Thus, a ∈ { 1 , 64 } . The only possibility is 133 = 64 + 64 + 1 + 1 + 1 + 1 + 1, so i i ( ) 7 7 | a | , . . . , | a | consists of 2 2’s and 5 1’s. It follows that the answer is · 2 = 2688 . 1 7 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND