Difference between revisions of "1985 OIM Problems/Problem 5"

Latest revision as of 00:50, 23 December 2023

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

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.

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 == Solution ==
-{{solution}}
+Using rule (i) and (iii): <math>f(10)=f(2)+f(5)=0</math>.  Since we need to assign a non-negative integer, then <math>f(2)=f(5)=0</math>
+Using rule (i): <math>f(1985)=f(5)+f(397)=f(397)</math>
+Using rule (iii) and (ii): <math>f(9)+f(397)=f(9*397)=f(3573)=0</math>
+Since we need to assign a non-negative integer, then <math>f(9)=f(397)=0</math>
+Therefore, <math>f(1985)=f(397)=0</math>
+~Tomas Diaz. ~orders@tomasdiaz.com
+{{Alternate solutions}}
 == See also ==
 https://www.oma.org.ar/enunciados/ibe1.htm