HMMT 二月 2023 · ALGNT 赛 · 第 6 题

HMMT February 2023 — ALGNT Round — Problem 6

Suppose a , a , . . . , a are positive real numbers such that 1 2 100 ka k − 1 a = k a − ( k − 1) k − 1 for k = 2, 3, . . . , 100. Given that a = a , compute a . 20 23 100

解析

Suppose a , a , . . . , a are positive real numbers such that 1 2 100 ka k − 1 a = k a − ( k − 1) k − 1 for k = 2, 3, . . . , 100. Given that a = a , compute a . 20 23 100 Proposed by: Sean Li, Vidur Jasuja Answer: 215 Solution: If we cross multiply, we obtain a a = na + ( n − 1) a , which we can rearrange and n n − 1 n − 1 n factor as ( a − n )( a − ( n − 1)) = n ( n − 1) . n n − 1 Let b = a − n . Then, b b = n ( n − 1). If we let b = t , then we have by induction that b = nt if n n n n − 1 1 n n is odd and b = n/t if n is even. So we have n { nt + n if n odd a = n n/t + n if n even for some real number t . We have 20 /t + 20 = 23 t + 23, so t ∈ {− 1 , 20 / 23 } . But if t = − 1, then a = 0 1 which is not positive, so t = 20 / 23 and a = 100 /t + 100 = 215. 100