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Algebra Difficulty 5.0 AIME Find the answer

Consider all functions f:ZZf: \mathbb{Z} \rightarrow \mathbb{Z} satisfying f(f(x)+2x+20)=15f(f(x)+2 x+20)=15 Call an integer nn good if f(n)f(n) can take any integer value. In other words, if we fix nn, for any integer mm, there exists a function ff such that f(n)=mf(n)=m. Find the sum of all good integers xx.

A number or a short expression. Spacing and $ signs are ignored.

Solution

For almost all integers x,f(x)x20x, f(x) \neq-x-20. If f(x)=x20f(x)=-x-20, then f(x20+2x+20)=15x20=15x=35f(-x-20+2 x+20)=15 \Longrightarrow-x-20=15 \Longrightarrow x=-35 Now it suffices to prove that the f(35)f(-35) can take any value. f(35)=15f(-35)=15 in the function f(x)15f(x) \equiv 15. Otherwise, set f(35)=cf(-35)=c, and f(x)=15f(x)=15 for all other xx. It is easy to check that these functions all work.

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