PUMaC 2019 · 代数（B 组） · 第 1 题

PUMaC 2019 — Algebra (Division B) — Problem 1

Let a, b be positive integers such that a + b = 10. Let be the difference between the maximum q 1 1 and minimum possible values of + , where p and q are relatively prime positive integers. a b Compute p + q. x 3 3 x

解析

Let a, b be positive integers such that a + b = 10. Let be the difference between the maximum q 1 1 and minimum possible values of + , where p and q are relatively prime. Compute p + q. a b Answer: 77 Proposed by: Matthew Kendall 1 1 1 1 10 10 Since b = 10 − a , + = + = = . For the maximum, we wish 2 a b a 10 − a a (10 − a ) − a +10 a to minimize the value of the denominator. That is achieved at the axis of symmetry of the 1 1 2 parabola, when a = 5, making + = . The minimum is achieved when the denominator is a b 5 maximized, which is when a is as far from the axis of symmetry as possible: a = 1 or a = 9. 1 1 1 1 10 Either value gives + = + = . a b 9 1 9 10 2 32 Hence, the difference of the minimum and maximum is − = , making the answer 9 5 45 32 + 45 = 77 . x 3 3 x