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Algebra Difficulty 4.8 AIME Prove it China

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

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

Solution

We determine f(2002)f(2002) first. From the conditions given, we have
f(x)+5≤f(x+5)≤f(x+4)+1≤f(x+3)+2≤f(x+2)+3≤f(x+1)+4≤f(x)+5. \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)+1f(x+1)=f(x)+1.

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

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