HMMT 二月 2002 · 冲刺赛 · 第 55 题
HMMT February 2002 — Guts Round — Problem 55
题目详情
- [10] A sequence of positive integers is given by a = 1 and a = gcd( a , n ) + 1 for 1 n n − 1 n > 1. Calculate a . 2002 2
解析
- A sequence of positive integers is given by a = 1 and a = gcd( a , n ) + 1 for 1 n n − 1 n > 1. Calculate a . 2002 Solution: 3 . It is readily seen by induction that a ≤ n for all n . On the other hand, n a is one greater than a divisor of 1999. Since 1999 is prime, we have a = 2 or 2000; 1999 1999 the latter is not possible since 2000 > 1999, so we have a = 2. Now we straightforwardly 1999 compute a = 3 , a = 4, and a = 3. 2000 2001 2002 2