1985 OIM Problems/Problem 5

Problem

To each positive integer $n$ we assign an integer non-negative $f(n)$ such that these conditions are satisfied:

(i) $f(rs)=f(r)+f(s)$

(ii) $f(n)=0$, when the unit digit of $n$ is 3

(iii) $f(10)=0$

Find $f(1985)$. Justify your answer.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Using rule (i) and (iii): $f(10)=f(2)+f(5)=0$. Since we need to assign a non-negative integer, then $f(2)=f(5)=0$

Using rule (i): $f(1985)=f(5)+f(397)=f(397)$

Using rule (iii) and (ii): $f(9)+f(397)=f(9*397)=f(3573)=0$

Since we need to assign a non-negative integer, then $f(9)=f(397)=0$

Therefore, $f(1985)=f(397)=0$


~Tomas Diaz. ~orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

https://www.oma.org.ar/enunciados/ibe1.htm