PUMaC 2010 · 数论（B 组） · 第 6 题
PUMaC 2010 — Number Theory (Division B) — Problem 6
题目详情
- Given that x , y are positive integers with x ( x + 1) | y ( y + 1), but neither x nor x + 1 divides 2 2 2 2 either of y or y + 1, and x + y as small as possible, find x + y .
解析
- Given that x , y are positive integers with x as small as possible, and y minimized with that constraint, and x ( x + 1) | y ( y + 1), but neither x nor x + 1 divides either of y or y + 1, find 2 2 x + y . Solution: x = 14, y = 20. These are the first two times that n and n + 1 are not powers of primes. The answer is then easily seen to be 596.