HMMT 二月 2020 · 冲刺赛 · 第 21 题

HMMT February 2020 — Guts Round — Problem 21

[12] We call a positive integer t good if there is a sequence a , a , . . . of positive integers satisfying 0 1 a = 15 , a = t , and 0 1 a a = ( a − 1)( a + 1) n − 1 n +1 n n for all positive integers n . Find the sum of all good numbers.

解析

[12] We call a positive integer t good if there is a sequence a , a , . . . of positive integers satisfying 0 1 a = 15 , a = t , and 0 1 a a = ( a − 1)( a + 1) n − 1 n +1 n n for all positive integers n . Find the sum of all good numbers. Proposed by: Krit Boonsiriseth Answer: 296 Solution: By the condition of the problem statement, we have 2 2 a − a a = 1 = a − a a . n − 1 n +1 n − 2 n n n − 1 This is equivalent to a + a a + a n − 2 n n − 1 n +1 = . a a n − 1 n a + a 0 2 Let k = . Then we have a 1 a + a a + a a + a a + a n − 1 n +1 n − 2 n n − 3 n − 1 0 2 = = = · · · = = k. a a a a n n − 1 n − 2 1 Therefore we have a = ka − a for all n ≥ 1. We know that k is a positive rational number n +1 n n − 1 because a , a , and a are all positive integers. We claim that k must be an integer. Suppose that 0 1 2 p k = with gcd( p, q ) = 1. Since ka = a + a is always an integer for n ≥ 1, we must have q | a n n − 1 n +1 n q 2 for all n ≥ 1. This contradicts a − a a = 1. Conversely, if k is an integer, inductively all a are 1 3 i 2 integers. 2 2 t − 1 t +224 224 Now we compute a = , so k = is an integer. Therefore 15 k − t = is an integer. 2 15 15 t t Combining with the condition that a is an integer limits the possible values of t to 1, 4, 14, 16, 56, 2
The values t < 15 all lead to a = 0 for some n whereas t > 15 leads to a good sequence. The n sum of the solutions is 16 + 56 + 224 = 296 .