PUMaC 2010 · 代数（B 组） · 第 6 题

PUMaC 2010 — Algebra (Division B) — Problem 6

Define f ( x ) = x + x + x + x + x + . . . . Find the smallest integer x such that f ( x ) ≥ √ 50 x . x

解析

Define f ( x ) = x + x + x + x + x + . . . . Find the smallest integral x such that f ( x ) ≥ √ 50 x . 2 Solution: 2400. Noting that ( f ( x ) − x ) = f ( x ), we can solve the quadratic equation for f ( x ) to get that √ 1 1 f ( x ) = x + ± x + . 2 4 We clearly have to take the positive root (we can notice, for example, that f (1) > 1). The problem therefore reduces to finding the smallest integral x such that √ √ 1 1 x + + x + ≥ 50 x. 2 4 It is simple to note that x has to be fairly large for this to be satisfied (after trying the trivial √ √ 1 x = 1). For large x , x + is very, very close to x , so we can rewrite this as 4 √ 1 x + ≥ 49 x. 2 1 The above is again rewritten as 1 2 x − 2400 x + ≥ 0 . 4 The smallest integer x satisfying the above is obviously 2400, and since the margin of error 1 here is , our previous approximation is justified. 4 Of note is that this problem is cooked: the value x = 0 is also a valid solution. We accepted either solution. x