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Problem 1127

AMC 12 late, AIME early

Algebra Difficulty 4.8 Prove it China Mathematical Competition · China

It is given that $f(x)$ is a function defined on $\mathbb{R}$ , satisfying $f(1) = 1$ , and for any $x \in \mathbb{R}$ ,
$f(x+5) \ge f(x)+5,$

and $f(x+1) \le f(x)+1$ .
If $g(x) = f(x)+1-x$ , then $g(2002) = \underline{\hspace{2cm}}$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

We determine $f(2002)$ first. From the conditions given, we have
$\begin{aligned} f(x)+5 &\le f(x+5) \le f(x+4)+1 \\ &\le f(x+3)+2 \le f(x+2)+3 \\ &\le f(x+1)+4 \le f(x)+5. \end{aligned}$
Thus the equality holds for all. So we have $f(x+1)=f(x)+1$ .

Hence, from $f(1)=1$ , we get $f(2)=2$ , $f(3)=3$ , ..., $f(2002)=2002$ . Therefore, $g(2002)=f(2002)+1-2002=1$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.