Difference between revisions of "1985 OIM Problems/Problem 5"
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Find <math>f(1985)</math>. Justify your answer. | Find <math>f(1985)</math>. Justify your answer. | ||
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| + | ~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ||
== Solution == | == Solution == | ||
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| + | 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> | ||
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| + | 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> | ||
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| + | Since we need to assign a non-negative integer, then <math>f(9)=f(397)=0</math> | ||
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| + | Therefore, <math>f(1985)=f(397)=0</math> | ||
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| + | ~Tomas Diaz. ~orders@tomasdiaz.com | ||
| + | |||
| + | {{Alternate solutions}} | ||
| + | |||
| + | == See also == | ||
| + | https://www.oma.org.ar/enunciados/ibe1.htm | ||
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.
