Solution

38915

First name
Iyad Mohamed
School
Garden International School
Country
Age
12

So we started by picking four numbers that couldn't in anyway, add up to a multiple of three with three out of four said numbers. Then we tried to find the fifth number, which was the hard part.

After some guessing and after many attemps we realized that it was impossible and that we had wasted the previous 20 minutes trying to do something that we feel is clearly not possible.

Before we did some trial and area, we thought of some rules that would be impossible to find in the number. They were:
That it couldn't have three of the same number.
That it couldn't have a multiple of three with both a positive and negative version of the same number. (Such as 0, 3, 5, 4, -4 because 3 (the multiple or three) and 4 and -4 (the positive and negative versions of itself) equals to a multiple of three (3))
And that you couldn't have a zero.

Our guesses were:
1 8 2 4
7 7 8 8
2 2 5 5
13 15 17 19
6 5 8 2
3 4 10 100
1 7 10 100
And 1 5 7 17

We chose to work on the numbers "1, 5, 7, 17".

For the numbers "1, 5, 7, 17", all our guesses were for the fifth number. We trialled our numbers by finding all the combinations of two numbers in "1, 5, 7, 17" which was "6, 8, 18, 12, 22 and 24" Our trialled numbers include:
1| 7, 9, 19, 13, 23, 25
2| 8, 10, 20, 14, 24, 26
4| 10, 12 (discontinued)
8| 14, 16, 26, 20, 30, 32
-2| 4, 6 (discontinued)
-4| 2, 4, 14, 8, 18, 20
-5| 1, 3 (discontinued)
20| 26, 28, 38, 32, 42, 44

So as you can see, there will always be at least one multiple of three in every five numbers.

Then we started to look at the remainders. If you have 3 numbers with a remainder of one it will all add up to a multiple of 3 the same goes with 3 numbers with a remainder of 2 it will add up to another multiple of 3. So then the first 4 numbers must be
Remainder 1, 1, 2, 2

We can't add another remainder 1 or 2 so we must add one that is a multiple of 3, remainder 0. However they can add up: remainder 1, 2 and 0 add up to remainder 3 or remainder 0 so no matter what you can't have 5 numbers which 3 of which can't add up to a number that is a multiple of 3.

It works a bit like this:

r^1, r^1, r^1, r^2, r^0
r^2, r^2, r^2, r^1, r^0

r^0, r^0, r^0, r^1, r^2

r^1, r^1, r^2, r^2, r^0