HMMT 二月 2011 · ALGCALC 赛 · 第 7 题
HMMT February 2011 — ALGCALC Round — Problem 7
题目详情
英文原题
- Let f : [0 , 1) → R be a function that satisfies the following condition: if
∞
∑
x = = .a a a . . .na
1 2 3
10 nn =1
is the decimal expansion of x and there does not exist a positive integer k such that a = 9 for allnn ≥ k , then
∞
∑
f ( x ) = .na
2 nn =110
( )
′1
Determine f .
23
解析
英文解析
- As a result, 2 x − 3 = 7 or − 7 for some positive integer n , or either x + 2 or 2 x − 3 is ± 1. Weconsider 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
• (2 x − 3) = ± 7 , for n ≥ 2. Then, since x + 2 = , we have that x + 2 is divisible by 7 nbut not by 49. Hence x + 2 = ± 7, yielding x = 5 , − 9. The former has already been considered,2
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 noticedon the day of the test, and only the answer − 3 , 2 , 5 was marked as correct.