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Take Three from Five

Stage: 3 and 4 Challenge Level: Challenge Level:2 Challenge Level:2
We had lots of good solutions to this problem, thanks to everyone who sent us theirs!
Felix from the German Swiss International School in Hong Kong sent us this solution:

It is easier to look at even numbers first.

All numbers are either even or odd. Even numbers can be written in the form $2n$, odd numbers can be written in the form $2n+1$

From common sense, we know that adding two even numbers is an even number. Proof:
$(2a) + (2b) = 2(a + b)$, which is a multiple of $2$

Also, if we add two odd numbers, we get another even number.
$(2a + 1) + (2b + 1) = 2(a + b + 1)$

We can always get an even number from the sum of $2$ selected numbers chosen from 3 random numbers, we can tell by looking at the possible cases:
  1. All three are even
  2. Two are even, one is odd
  3. One is even, two are odd
  4. All three are odd
  In each case we can choose two numbers with the same form, both even or both odd.

Now, using a similar method above, we can prove that we can always get a multiple of $3$ from the sum of $3$ selected numbers chosen from any $5$ numbers.

Since we are talking about multiples of $3$, we will say that all the whole numbers are either a multiple of three (in the form $3n$), one more than a multiple of $3$ (in the form $3n + 1$), or two more than a multiple of $3$ (in the form $3n + 2$).

We can get a multiple of $3$ if we add three multiples of three.
$3a + 3b + 3c = 3(a + b + c)$

We can get a multiple of $3$ if we add $3$ numbers in the form $3n + 1$.
$(3a + 1) + (3b + 1) + (3c + 1) = 3a + 3b + 3c + 3 = 3(a + b + c + 1)$

We can get a multiple of $3$ if we add $3$ numbers in the form $3n + 2$. Proof:

$(3a + 2) + (3b + 2) + (3c + 2) = 3a + 3b + 3c + 6 = 3(a + b + c + 2)$

So if we wanted to find a set of $5$ numbers where we couldn't find $3$ which added up to a multiple of $3$, we would want to avoid five numbers containing $3$ of the same form.

If we don't want $3$ of the same form we need $2$ of one form, $2$ another form and $1$ of the last form.

This has at least one of each form.

If we take one from each form we get a multiple of $3$

$3a + (3b + 1) + (3c + 2) = 3a + 3b + 3c + 3 = 3(a + b + c + 1)$

ANSWER: It is always possible to get a multiple of $3$ adding $3$ numbers from any $5$ random numbers by selecting $3$ numbers of the same form ($3n$, $3n+ 1$, or $3n + 2$) or by selecting $3$ numbers with different forms ($3n$, $3n + 1$ and $3n + 2$), and its always possible to do this.

We also received a very colourful solution from Marie and Caitlin from St Leonard's Catholic School, which you can see here.
Well done!