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!

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!