PUMaC 2020 · 数论（B 组） · 第 1 题
PUMaC 2020 — Number Theory (Division B) — Problem 1
题目详情
- The number 2021 leaves a remainder of 11 when divided by a positive integer. Find the smallest such integer.
解析
- The number 2021 leaves a remainder of 11 when divided by a positive integer. Find the smallest such integer. Proposed by: Frank Lu Answer: 15 Let n be our number. Then, we need our number to divide 2021 − 11 = 2010 , and to be strictly greater than 11 . However, we know that 2010 = 2 · 3 · 5 · 67 . In other words, we are looking for the smallest divisor of 2010 that is larger than 11 . We see that this is 15 .