HMMT 二月 2006 · 冲刺赛 · 第 5 题
HMMT February 2006 — Guts Round — Problem 5
题目详情
- [6] Find the number of solutions in positive integers ( k ; a , a , . . . , a ; b , b , . . . , b ) to the 1 2 1 2 k k equation a ( b ) + a ( b + b ) + · · · + a ( b + b + · · · + b ) = 7 . 1 1 2 1 2 k 1 2 k
解析
- Find the number of solutions in positive integers ( k ; a , a , . . . , a ; b , b , . . . , b ) to the 1 2 k 1 2 k equation a ( b ) + a ( b + b ) + · · · + a ( b + b + · · · + b ) = 7 . 1 1 2 1 2 k 1 2 k Answer: 15 Solution: Let k , a , . . . , a , b , . . . , b be a solution. Then b , b + b , . . . , b + · · · + b is 1 k 1 k 1 1 2 1 k just some increasing sequence of positive integers. Considering the a as multiplicities, i the a ’s and b ’s uniquely determine a partition of 7. Likewise, we can determine a ’s i i i and b ’s from any partition of 7, so the number of solutions is p (7) = 15 . i