HMMT 二月 2000 · ADV 赛 · 第 4 题
HMMT February 2000 — ADV Round — Problem 4
题目详情
英文原题
- Fiv e p ositiv e in tegers from 1 to 15 are hosen without repla emen t. What is the probabilit y that their sum is divisible b y 3?
解析
英文解析
- The p ossibilities for the n um b ers are:
all v e are divisible b y 3
three are divisible b y 3, one is 1 (mo d 3) and one is 2 (mo d 3)
t w o are divisible b y 3, and the other three are either 1 (mo d 3) or 2 (mo d 3)
one is divisible b y 3, t w o are 1 (mo d 3) and t w o are 2 (mo d 3)
four are 1 (mo d 3) and one is 2 (mo d 3)
four are 2 (mo d 3) and one is 1 (mo d 3)
This giv es us 1001 p ossible om binations out of or 3003. So, the probabilit y is 15
100115 = .
3003 3