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The group at Henry Cavendish Primary School thought they found 5 numbers that worked - 6, 17, 5, 41, 96 - and asked for some feedback! This was a really good try, but unfortunately there is a multiple of three in there: 17 + 5 + 41 is equal to 3 times 21. Never mind!

The answer is that it's not possible. Lots of people got this, but David, from Sha Tin College, gave a particularly well written answer:


Firstly, each one of the five integers can be expressed in the form 3x + 0, 3x + 1 or 3x + 2, where x is an integer.

When we take three integers (say 3x + a, 3y + b, 3z + c) from a group 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 a, b and c)

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 a, b and c are each equal to 0, 1 or 2, there are 4 different ways in which k can be a multiple of 3:

1. 0 + 0 + 0 (i.e. a=0, b=0, c=0)
2. 1 + 1 + 1
3. 2 + 2 + 2
4. 0 + 1 + 2

Out of a group of 5 integers, it is always possible to find an a, b and c that 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.

The Year 5 group at St. John's C of E Primary remarked that it is possible to make a set of 5 numbers like this if you don't insist that they have to be whole numbers: they came up with 0.2, 0.2, 0, 0, 1. That's definitely thinking outside the box!

Thanks to everyone - lots of clever answers here!