HMMT 十一月 2019 · GEN 赛 · 第 8 题

HMMT November 2019 — GEN Round — Problem 8

Compute the number of ordered pairs of integers ( x, y ) such that x + y < 2019 and 2 2 x + min( x, y ) = y + max( x, y ) .

解析

Compute the number of ordered pairs of integers ( x, y ) such that x + y < 2019 and 2 2 x + min( x, y ) = y + max( x, y ) . 2 Proposed by: Milan Haiman Answer: 127 We have 2 2 x − y = max( x, y ) − min( x, y ) = | x − y | Now if x 6 = y , we can divide by x − y to obtain x + y = ± 1. Thus x = y or x + y = ± 1. 2 2 2 If x = y , we see that 2019 > x + y = 2 x , so we see that − 31 ≤ x ≤ 31. There are 63 ordered pairs in this case. 2 2 In the second case, note that | x | ≥ | y | since x − y = | x − y | ≥ 0. Since x + y = ± 1, we cannot have xy > 0, so either x ≥ 0 , y ≤ 0, or x ≤ 0 , y ≥ 0. In the first case, x + y = 1; in the second case, x + y = − 1. Thus, the solutions for ( x, y ) are of the form ( k, 1 − k ) or ( − k, k − 1) for some k > 0. In either case, we 2 2 must have k + ( k − 1) < 2019, which holds true for any 1 ≤ k ≤ 32 but fails for k = 33. There are a total of 32 · 2 = 64 solutions in this case. 2 In summary, there are a total of 63 + 64 = 127 integer solutions to the equation x + min ( x, y ) = 2 2 2 y + max ( x, y ) with x + y < 2019.