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
3x + c, where x is an integer and c is either 0, 1 or 2. When
we take three integers (say 3x +c1, 3y + c2, 3z + c3) from
a set of 5 and add them together, we will always get an integer in
the following form:
which equals
(where k is the sum of c1, c2 and c3)
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.