Competition problem — Number theory

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Number theory Difficulty 4.5 AIME Find the answer

The function $\mathrm{f}(\mathrm{n})$ is defined on the positive integers and takes non-negative integer values. It satisfies (1) $f(m n)=f(m)+f(n),(2) f(n)=0$ if the last digit of $n$ is 3, (3) $f(10)=0$ . Find $\mathrm{f}(1985)$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

If $\mathrm{f}(\mathrm{mn})=0$ , then $\mathrm{f}(\mathrm{m})+\mathrm{f}(\mathrm{n})=0$ (by $(1))$ . But $\mathrm{f}(\mathrm{m})$ and $\mathrm{f}(\mathrm{n})$ are non-negative, so $\mathrm{f}(\mathrm{m})=\mathrm{f}(\mathrm{n})=0$ . Thus $f(10)=0$ implies $f(5)=0$ . Similarly $f(3573)=0$ by (2), so $f(397)=0$ . Hence $f(1985)$ $=\mathrm{f}(5)+\mathrm{f}(397)=0$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.