Take Three from Five

Firstly, each one of the five integers can be expressed in the form $3x + c$, where $x$ is an integer and $c$ is either 0, 1 or 2. When we take three integers (say $3x +c_1$, $3y + c_2$, $3z + c_3$) from a set of 5 and add them together, we will always get an integer in the following form: $$3x + 3y + 3z + k,$$ 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.