Please prove the following statement:
In every set of 5 integers one can always select 3 members such that their sum is divisible by 3.
Source: Crux Mathematicorum
REM: The original puzzle specifies positive integers. Considering this word as redundant I have erased it.
Do you agree?
Assume the statement is false.
By definition, every number is worth either 0, 1, or 2, mod3.
Assume we select one of each type: 0+1+2=6, worth 0, mod 3.
So there cannot be one of each type among the 5. Then there must be at most two types, and since there are 5 selections, one type must have at least 3 examples.
But if so, then either:
0+0+0=0, worth 0, mod3.
or
1+1+1=3, worth 0, mod3.
or
2+2+2=6, worth 0, mod3.
This is a contradiction, so the statement must be true.
Edited on July 16, 2022, 9:30 pm
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Posted by broll
on 2022-07-16 07:29:06 |
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