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> | ||
+ | |||
+ | <math>f(1985)=f(5)+f(397)=f(397)</math> | ||
+ | |||
+ | <math>f(9)+f(397)=f(9*397)=f(3573)=0 | ||
+ | |||
+ | 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> | ||
+ | |||
+ | |||
+ | ~Tomas Diaz. ~orders@tomasdiaz.com | ||
+ | |||
+ | {{Alternate solutions}} | ||
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe1.htm | https://www.oma.org.ar/enunciados/ibe1.htm |
Revision as of 00:49, 23 December 2023
Problem
To each positive integer we assign an integer non-negative such that these conditions are satisfied:
(i)
(ii) , when the unit digit of is 3
(iii)
Find . Justify your answer.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Since we need to assign a non-negative integer, then
$f(9)+f(397)=f(9*397)=f(3573)=0
Since we need to assign a non-negative integer, then$ (Error compiling LaTeX. Unknown error_msg)
Therefore,
~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.