This problem has been solved by Felix from the German Swiss International School, Hong Kong, Curt from Reigate College, Tom who does not tell us his school and David from Sha Tin College, also in Hong Kong.

Here is David's solution:

Firstly, each one of the five integers can be expressed in the form

3 x + c

, where

x

is an integer and

c

is either 0, 1 or 2. When we take three integers (say

3 x + c_{1}

3 y + c_{2}

3 z + c_{3}

) from a set of 5 and add them together, we will always get an integer in the following form:

3 x + 3 y + 3 z + k,

which equals

3 (x + y + z) + k

(where k is the sum of

c_{1}

c_{2}

and

c_{3}

)

It is obvious that the term 3(x + y + z) must be a multiple of 3. Therefore, we only need to consider $k$ in order to see whether $3 (x + y + z) + k$ is a multiple of 3.

For $3 (x + y + z) + k$ to be a multiple of 3, $k$ must therefore also be a multiple of 3.

As $c = 0, 1$ or 2 therefore there are 4 different ways in which $k$ can be a multiple of 3:

1. 0 + 0 + 0
2. 1 + 1 + 1
3. 2 + 2 + 2
4. 0 + 1 + 2

Out of a group of 5 integers, it is always possible for three values of $c$ to add up to a multiple of 3. This is because in a group of 5 zeros, ones or twos, there is either at least one of each number (0,1,2), or at least 3 of just one of the numbers (0,0,0 or 1,1,1 or 2,2,2).

Therefore it is always possible to choose three numbers that will add up to a multiple of 3 from any group of 5 numbers.