HMMT 二月 2001 · ADV 赛 · 第 8 题

英文解析

8. Find the number of positive integer solutions to n + n = n with n < 2001.
   Solution: If n = 1, the relation can not hold, so assume otherwise. If x > y , the lefty x − y x − yhand side factors as n ( n + 1) so n + 1 is a power of n . But it leaves a remainder of 1
   when divided by n and is greater than 1, a contradiction. We reach a similar contradictionx zif y > x . So y = x and 2 n = n , so 2 is a power of n and n = 2. So all solutions are of thex x x +1 x +1
   form 2 + 2 = 2 , which holds for all x . 2 < 2001 implies x < 11, so there are 10
   solutions.