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I start with a red, a green and a blue marble. I can trade any of my marbles for two others, one of each colour. Can I end up with five more blue marbles than red after a number of such trades?

More Marbles

I start with a red, a blue, a green and a yellow marble. I can trade any of my marbles for three others, one of each colour. Can I end up with exactly two marbles of each colour?


Can you guarantee that, for any three numbers you choose, the product of their differences will always be an even number?


Age 11 to 14
Challenge Level
We had loads of great submissions for this problem - thank you to everyone for thinking so creatively and having some great answers!

Charlie from Fortismere School explains why Sinead doesn't have enough coins to solve her problem:

I did the maths practically using buttons and putting them in $10$ bowls.I used 44 buttons and tried to put them into the bowls so that each bowl had a different number of buttons.That didn't work so I used $45$ buttons instead and tried to put them in the ten bowls. I decided that one of the bowls
could have no buttons in it. I found that this time I was successful.

So then I did the maths on paper and saw that $0-9$ added up to $45$. $45$ is the
minimum number of coins Sinead can have in $10$ pockets because $0-9$ is the
lowest set of ten different positive numbers that we can have.

Mitchell from St. Mary and All Saints Primary School gave a great explanation for why Abbie's problem is impossible:

I know this is impossible because if you add $1$,$2$,$3$,$4$,$5$,$6$,$7$,$8$,$9$, and $10$ together you will find that it equals $55$. $55$ is an odd number so if you halve it,  it will not be a whole number. Therefore if Abbie wants $2$ towers split evenly, she will need to split one of the cubes
into half.

Max from Hitchin Boy's School goes on to describe which numbers Abbie would be able to build two identical towers for:

The two towers will have a total height of $\sum{n} = 1+ 2 + ... + n = \frac{1}{2}n(n+1)$, where n is the number of cubes Abbie has.
If $\frac{1}{2}n(n+1)$ is odd then it can't be divided into 2 equal towers.
$\frac{1}{2}n(n+1)$ will be even if n is a multiple of $4$ or one less than a multiple of $4$.

Erin from Hollyfield used algebra to explain why the sum of four consecutive numbers can never be a multiple of $4$:

Let's call our starting number $n$.
$(n)+(n+1)+(n+2)+(n+3)=4n+6$. This means that out first number $n$ will be multiplied by 4, but then 6 will be added. Because a multiple of $4$ plus $6$ can never make a multiple of $4$, this means that the numbers in the sequence added up will never be a multiple
of $4$. 

Thank you everyone for all of your solutions - keep being creative!