PUMaC 2012 · 代数（A 组） · 第 1 题

PUMaC 2012 — Algebra (Division A) — Problem 1

[ 3 ] Compute the smallest positive integer a for which a + a + ... − > 7. 1 a + a + .. x − y y − z z − x

解析

[ 3 ] Compute the smallest positive integer a for which a + a + ... − > 7. 1 a + a + .. √ 1 Solution: Let the first term be x and the second y . Then we have x = a + x and y = a + y After solving this system in terms of a , we can write the difference as √ √ 1 2 ( a + 1 + 4 a + 1 − a + 4) 2 Further simplification gives: √ √ 2 a − a + 4 + 4 a + 1 > 13 , where the first term is a very small negative number. Thus a = 42 does not quite get us there but a = 43 does. Answer: 43 x − y y − z z − x