HMMT 二月 2011 · ALGCALC 赛 · 第 7 题

HMMT February 2011 — ALGCALC Round — Problem 7

Let f : [0 , 1) → R be a function that satisfies the following condition: if ∞ ∑ a n x = = .a a a . . . 1 2 3 n 10 n =1 is the decimal expansion of x and there does not exist a positive integer k such that a = 9 for all n n ≥ k , then ∞ ∑ a n f ( x ) = . 2 n 10 n =1 ( ) 1 ′ Determine f . 3 2

解析

As a result, 2 x − 3 = 7 or − 7 for some positive integer n , or either x + 2 or 2 x − 3 is ± 1. We consider the following cases: • (2 x − 3) = 1. Then x = 2, which yields f ( x ) = 4, a prime power. • (2 x − 3) = − 1. Then x = 1, which yields f ( x ) = − 3, not a prime power. • ( x + 2) = 1) . Then x = − 1, which yields f ( x ) = − 5 not a prime power. • ( x + 2) = − 1. Then x = − 3, which yields f ( x ) = 9, a prime power. • (2 x − 3) = 7. Then x = 5, which yields f ( x ) = 49, a prime power. • (2 x − 3) = − 7. Then x = − 2, which yields f ( x ) = 0, not a prime power. (2 x − 3) + 7 n • (2 x − 3) = ± 7 , for n ≥ 2. Then, since x + 2 = , we have that x + 2 is divisible by 7 2 but not by 49. Hence x + 2 = ± 7, yielding x = 5 , − 9. The former has already been considered, while the latter yields f ( x ) = 147. So x can be either -3, 2 or 5. (Note: In the official solutions packet we did not list the answer -3. This oversight was quickly noticed on the day of the test, and only the answer − 3 , 2 , 5 was marked as correct.