HMMT 二月 2020 · ALGNT 赛 · 第 2 题

HMMT February 2020 — ALGNT Round — Problem 2

Find the unique pair of positive integers ( a, b ) with a < b for which 2020 − a 2020 − b · = 2 . a b 2

解析

Find the unique pair of positive integers ( a, b ) with a < b for which 2020 − a 2020 − b · = 2 . a b Proposed by: James Lin Answer: (505 , 1212) Solution 1: If either a or b is larger than 2020, then both must be for the product to be positive. However, the resulting product would be less than 1, so this case is impossible. Now, we see that 2 y 2020 − a 2020 − b x ( , ) must be in the form ( , ), in some order, for relatively prime positive integers x and a b y x x + y x +2 y 2020 2020 y . Then = and = , so x + y and x + 2 y are relatively prime factors of 2020. Since a y b x x + y < x + 2 y < 2( x + y ), the only possibility is x + y = 4, x + 2 y = 5. Thus, ( x, y ) = (3 , 1), and 2020 − a 2020 − b 2 ( , ) = (3 , ) because a < b . Solving gives ( a, b ) = (505 , 1212). a b 3 2 Solution 2: We rearrange to find that ( a +2020)( b +2020) = 2 · 2020 . Note that a +2020 and b +2020 2 are both less than 4040 < 101 , so they must both be divisible by 101. Hence, we divide out a factor 2 ′ ′ ′ a ′ b of 101 and solve the equivalent problem of ( a + 20)( b + 20) = 800, where a = and b = . 101 101 4040 ′ ′ Because each factor must be less than = 40, we get that ( a + 20 , b + 20) = (25 , 32), which yields 101 ( a, b ) = (505 , 1212). 2