HMMT 十一月 2015 · GEN 赛 · 第 6 题
HMMT November 2015 — GEN Round — Problem 6
题目详情
英文原题
- Consider all functions f : Z → Z satisfyingf ( f ( x ) + 2 x + 20) = 15 .
Call an integer n good if f ( n ) can take any integer value. In other words, if we fix n , for any integerm , there exists a function f such that f ( n ) = m. Find the sum of all good integers x.
解析
英文解析
- Consider all functions f : Z → Z satisfyingf ( f ( x ) + 2 x + 20) = 15 .
Call an integer n good if f ( n ) can take any integer value. In other words, if we fix n , for any integerm , there exists a function f such that f ( n ) = m. Find the sum of all good integers x.
Proposed by: Yang Liu
Answer: -35
For almost all integers x , f ( x ) 6 = − x − 20 . If f ( x ) = − x − 20 , thenf ( − x − 20 + 2 x + 20) = 15 = ⇒ − x − 20 = 15 = ⇒ x = − 35 .
Now it suffices to prove that the f ( − 35) can take any value.
f ( − 35) = 15 in the function f ( x ) ≡ 15 . Otherwise, set f ( − 35) = c , and f ( x ) = 15 for all other x . Itis easy to check that these functions all work.