A problem from the February 2006 issue of Crux Mathematicorum:
In a set of five positive integers, show that it’s always possible to choose three such that their sum is a multiple of 3.

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This solution is by Geneviève Lalonde. Any integer must fall into one of three groups:
 It’s a multiple of 3.
 It’s one less than a multiple of 3.
 It’s one more than a multiple of 3.
In our set of five numbers, either it’s the case that all three of these groups are represented or it’s not.
If all three are represented, then the set includes numbers of the form 3a – 1, 3b, and 3c + 1. The sum of these is 3a – 1 + 3b + 3c + 1 = 3(a + b + c), which is a multiple of 3.
If not all three groups are represented, then one of the groups must contain at least three numbers from our set of five.
 If there are three numbers in the first group, they’ll have the form 3a, 3b, and 3c.
 If there are three numbers in the second group, they’ll have the form 3a – 1, 3b – 1, and 3c – 1.
 If there are three numbers in the third group, they’ll have the form 3a + 1, 3b + 1, and 3c + 1.
In each case, the sum of the three numbers is a multiple of 3.
