Ben's Game

We received several responses mentioning that a strategy of trial and error had been used to arrive at the result. This is a valuable strategy but it may be difficult to tell if there is more than one solution.

Alice used a spreadsheet to help her consider the many possibilities :

First of all I considered what number of counters each of them could have.

Ben's have to be a multiple of $3$, but not $3$, Emma's have to go by $5$, but not $5$, and Jacks had to go by $4$ but not be $4$.

I decided to do a spread-sheet:
Ben has $2/3$ of his left and $1/5$ of Emma's, ($2B/3 + E/5$),
Jack has $3/4$ of his left and $1/3$ of Ben's ($3J/4 + B/3$),
and Emma has $4/5$ of hers left and $1/4$ of Jack's ($4E/5 + J/4$).

After a few goes at putting numbers into the spreadsheet, the answer turned out to be $12$ for Ben, $8$ for Jack and $10$ for Emily.

Zak and Sam from Norwich School for boys showed that this solution works:

Ian from Myton School reasoned as follows:

Ben's approach confirmed that there is a single solution:

$E$ is a multiple of $5$, $B$ of $3$ and $J$ of $4$.

$E$ can start with $10, 15, 20\ldots$ etc (starts at $10$ because cannot pass just $1$ counter)
If we use $15$ then passing $1/5$ (ie.$3$) leaves $12$. We cannot receive $1$ counter on its own so the finishing total for $E$ would be greater than $13$ which is not possible as this would require more than $40$ counters altogether. Therefore $E$ must have started with $10$.

If $E$ starts with $10$ then $B + J < 30$.
So $B$ can start with $6, 9, 12, 15, 18, 21\ldots$
and $J$ with $8, 12, 16, 20\ldots$

Only way to get $E, B$ and $J$ to finish with $13$ (the maximum possible),
is if $J = 20$ and $B = 9$,
or if $J = 8$ and $B = 21$
However these do not work, so $E, B$ and $J$ must finish with less than $13$.

Therefore we can eliminate other combinations leaving the options of
$B$ starting with either $6, 9, 12$ or $15$,
and $J$ starting with either $8, 12$ or $16$.

If $E = 10$,
$B + J = 14, 17, 20, 23, 26$ (since the total must be a multiple of $3$)
so try $6 + 8.$
This doesn't work so elimate 6 since it doesn't contribute to any other total.

Try $9 + 8$.
This doesn't work so elimate 9.

Therefore $B$ starts with either $12$ or $15$,
and $J$ starts with either $8$, $12$ or $16$.

$12 + 12 = 24$ and $12 + 15 = 27$, so elimate $J$ starting with $12$.

Therefore $B$ starts with either $12$ or $15$,
and $J$ starts with either $8$ or $16$.

$12 + 16 = 28$ and $15 + 16 = 31$, so elimate $J$ starting with $16$.

Therefore $B$ starts with either $12$ or $15$,
and $J$ starts with $8$.

But $B = 15$ and $J = 8$ does not work

so $B = 12$ and $J = 8$
$E= 10, B = 12, J = 8$

All finish with $10$ therefore use $30$ counters.

A student from Carres Grammar School used simultaneous equations to arrive at the solution:

Knowing that $x$ is a multiple of $3$, $y$ is a multiple of $4$ and $z$ is a multiple of $5$ then leads to the solution.

Well done to you all.