HMMT 二月 2000 · ADV 赛 · 第 4 题

HMMT February 2000 — ADV Round — Problem 4

Fiv e p ositiv e in tegers from 1 to 15 are hosen without repla emen t. What is the prob- abilit y that their sum is divisible b y 3?

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) 15 This giv es us 1001 p ossible om binations out of or 3003. So, the probabilit y is 5 1 1001 = . 3003 3