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

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== Solution ==
 
== Solution ==
  
<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) 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>
  
<math>f(1985)=f(5)+f(397)=f(397)</math>
+
Using rule (i): <math>f(1985)=f(5)+f(397)=f(397)</math>
  
<math>f(9)+f(397)=f(9*397)=f(3573)=0
+
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><math>f(9)=f(397)=0</math>
+
Since we need to assign a non-negative integer, then <math></math>f(9)=f(397)=0<math>
  
Therefore, <math>f(1985)=f(397)=0</math>
+
Therefore, </math>f(1985)=f(397)=0$
  
  

Revision as of 00:50, 23 December 2023

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 $$ (Error compiling LaTeX. Unknown error_msg)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