Nice problem that had my Year 5 and 6 pupils going for quite some time.
They struggled, as I expected, with the language to explain.
We eventually got to something like:
If all five numbers are multiples of three the any three added together is a multiple of three.
If four of them are multiples the pick three of those.
If none, one or two are multiples you have to use the others carefully. It's about the remainders. Three numbers with remainder 1 each will total a multiple of 3 as the three remainders make an extra 3!
When there is a mixture of remainder 1 and remainder 2 numbers you just have to get the right mix - a multiple of three with a remainder of each type, or three numbers with remainder 2. You can't have one remainder of one type with two of the other - you get either 4 or 5 extra which do not divide exactly by 3.
This was from a group who only recently saw that all the six times table answers are even.