Difference between revisions of "1970 Canadian MO Problems/Problem 7"
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== Solution == | == Solution == | ||
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| + | Let <math>a</math>, <math>b</math>, and <math>c</math> be three integers with equal modularity. | ||
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| + | That is, <math>a \equiv b\;(mod\;3)\equiv c\;(mod\;3)\equiv k\;(mod\;3)</math>, where <math>k=-1,0,</math> or <math>1</math>. | ||
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| + | In order for the sum of three integers to be divisible by three these three integers should have either all with the same modularity to each other, or they must have distinct modularity. | ||
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| + | ~Tomas Diaz. orders@tomasdiaz.com | ||
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| + | {{alternate solutions}} | ||
Revision as of 20:09, 27 November 2023
Problem
Show that from any five integers, not necessarily distinct, one can always choose three of these integers whose sum is divisible by 
.
Solution
Let 
, 
, and 
be three integers with equal modularity.
That is, 
, where 
or 
.
In order for the sum of three integers to be divisible by three these three integers should have either all with the same modularity to each other, or they must have distinct modularity.
~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.
