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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.

Even numbers are a multiple of $2$. If a number is not a multiple of $2$, then it is one more than a multiple of $2$, or what we call an odd number.

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

Also, if we add two odd numbers in the form $2n + 1$, we get another even number.
Proof:

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

We can always get an even number from the sum of $2$ selected numbers chosen from 3 random numbers because $2$ of the numbers must be both multiples of $2$, $2n$, or odd numbers, $2n + 1$. We can choose two numbers with the same form: $2n$ or $2n + 1$.

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 $5$ random 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.
Proof:
$3a + 3b + 3c = 3(a + b + c)$

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

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

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

The above is only if there are $3$ numbers of the same form. What if we don't get $3$ of the same form? Then we must have $2$ of one form, $2$ of another form, $1$ of the last form.

$2$ of one form and $1$ of another form does NOT work because the constants ($0$ for $3n$, $1$ and $2$) at the end will never add to $3$. For example:

$3a + 3b + (3c + 1) = 3(a + b + c) + 1$ - this is 1 more than a multiple of $3$ as $0 + 0 + 1 = 1$.

How can we add these different constants so that it is $3$ or a multiple of $3$? Take one from each form!
Proof:
$3a + (3b + 1) + (3c + 2) = 3a + 3b + 3c + 3 = 3(a + b + c + 1)$

ANSWER: It is possible to get a multiple of $3$ adding $3$ numbers from $5$ random numbers provided that you select $3$ numbers of the same form ($3n$, $3n+ 1$, or $3n + 2$) or if you select $3$ numbers with different forms ($3n$, $3n + 1$ and $3n + 2$).

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