This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 3

Consider a function $f: \mathbb Z \to \mathbb Z$ such that for every integer $n \ge 0$, there are at most $0.001n^2$ pairs of integers $(x,y)$ for which $f(x+y) \neq f(x)+f(y)$ and $\max\{ \lvert x \rvert, \lvert y \rvert \} \le n$. Is it possible that for some integer $n \ge 0$, there are more than $n$ integers $a$ such that $f(a) \neq a \cdot f(1)$ and $\lvert a \rvert \le n$? [i]Proposed by David Yang[/i]
Consider a function $f: \mathbb Z \to \mathbb Z$ such that for every integer $n \ge 0$, there are at most $0.001n^2$ pairs of integers $(x,y)$ for which $f(x+y) \neq f(x)+f(y)$ and $\max\{ \lvert x \rvert, \lvert y \rvert \} \le n$. Is it possible that for some integer $n \ge 0$, there are more than $n$ integers $a$ such that $f(a) \neq a \cdot f(1)$ and $\lvert a \rvert \le n$? [i]Proposed by David Yang[/i]
Consider a function $f: \mathbb Z \to \mathbb Z$ such that for every integer $n \ge 0$, there are at most $0.001n^2$ pairs of integers $(x,y)$ for which $f(x+y) \neq f(x)+f(y)$ and $\max\{ \lvert x \rvert, \lvert y \rvert \} \le n$. Is it possible that for some integer $n \ge 0$, there are more than $n$ integers $a$ such that $f(a) \neq a \cdot f(1)$ and $\lvert a \rvert \le n$? [i]Proposed by David Yang[/i]