Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Prove it Ibero-American Mathematical Olympiad

Problem:

The function $f(n)$ 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 $f(1985)$ .

Solution

Solution:

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

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