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3 out of 5 (Posted on 2022-07-15) Difficulty: 2 of 5
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?

No Solution Yet Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

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Solution Possible Solution Comment 2 of 2 |
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
  Posted by broll on 2022-07-16 07:29:06

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