HMMT 二月 2007 · 冲刺赛 · 第 27 题
HMMT February 2007 — Guts Round — Problem 27
题目详情
- [ 12 ] Find the number of 7-tuples ( n , . . . , n ) of integers such that
1 7
∑7
n = 96957 .6
i =1 i
10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUNDth
英文原题
[ 12 ] Find the number of 7-tuples ( n 1 , . . . , n 7 ) of integers such that
7 ∑
i =1
n 6
i = 96957 .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND
解析
英文解析
- [ 12 ] Find the number of 7-tuples ( n , . . . , n ) of integers such that
1 7
∑7
n = 96957 .6
i =1 i
Answer: 2688 . Consider the equation in modulo 9. All perfect 6 th 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 byi i i
6 6 6
3 , we have a + · · · + a = 133 . Since sixth powers are nonnegative, | a | ≤ 2. Again considering moduloi
1 7
9, we see that a 6 = 0. Thus, a ∈ { 1 , 64 } . The only possibility is 133 = 64 + 64 + 1 + 1 + 1 + 1 + 1, so 6
( )ii
| a | , . . . , | a | consists of 2 2’s and 5 1’s. It follows that the answer is · 2 = 2688 .77
1 7
10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUNDth