PUMaC 2012 · 数论（B 组） · 第 4 题

PUMaC 2012 — Number Theory (Division B) — Problem 4

[ 4 ] Albert has a very large bag of candies and he wants to share all of it with his friends. At first, he splits the candies evenly amongst his 20 friends and himself and he finds that there are five left over. Ante arrives, and they redistribute the candies evenly again. This time, there are three left over. If the bag contains over 500 candies, what is the fewest number of candies the bag can contain? 2012

解析

[ 4 ] Albert has a very large bag of candies and he wants to share all of it with his friends. At first, he splits the candies evenly amongst his 20 friends and himself and he finds that there are five left over. Ante arrives, and they redistribute the candies evenly again. This time, there are three left over. If the bag contains over 500 candies, what is the fewest number of candies the bag can contain? Solution: Note that 21 and 22 are relatively prime, so we can apply the Chinese Remainder Theorem to find that there is a unique solution modulo 462. Taking x to be the answer, we have x ≡ 5 (mod 21) x ≡ 3 (mod 22) Solving this system, we find that x ≡ 47 (mod 462) , so we have x = 509 . 2012