Difference between revisions of "1985 OIM Problems/Problem 5"
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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) 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 (i): <math>f(1985)=f(5)+f(397)=f(397)</math> | ||
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Using rule (iii) and (ii): <math>f(9)+f(397)=f(9*397)=f(3573)=0</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 < | + | Since we need to assign a non-negative integer, then <math>f(9)=f(397)=0</math> |
− | Therefore, < | + | Therefore, <math>f(1985)=f(397)=0</math> |
Latest revision as of 00:50, 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
Using rule (i) and (iii): . Since we need to assign a non-negative integer, then
Using rule (i):
Using rule (iii) and (ii):
Since we need to assign a non-negative integer, then
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.