# Ben's Game

*Ben's Game printable worksheet*

Ben, Jack and Emma were playing a game with a box of $40$ counters - they were not using all of them.

They each had a small pile of counters in front of them.

All at the same time, Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma, and Emma passed a fifth of her counters to Ben.

They all passed on more than one counter.

After this they all had the same number of counters.

**How many could each of them have started with?**

The total number they were playing with must be divisible by $3$.

The number of Ben's counters must initially be divisible by $3$, Jack's by $4$ and Emma's by $5$.

It might help to work out the maximum each could have started with -

e.g. Emma could not have started with $25$ counters. Can you work out why?

How many counters could each of them have started with?

Try some possible numbers.

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:

**(A)**$20x + 30y = 48z$ (from the first two equations)

**(B)**$15y + 36z = 40x$ (from the last two equations)

**(C)**$12z + 20x = 45y$ (from the first and last equations)

**(from A)**

**(D)**$ 40x = 96z - 60y$

**(from B and D)**

**(E)**$45y = 36z$

**(from C and E)**

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.

### Why do this problem?

This problem requires some simple knowledge of fractions and multiples and demands some strategic thinking. It may offer a good opportunity to compare methods between students - there isn't just one route to the solution. Note that there is no need to use algebra in this problem.

### Possible approach

*This printable worksheet may be useful: Ben's Game.*

Choose three students to act out the scenario with a (real or imaginary) pot of $40$ counters, as described under support below.

Ask students to work in pairs or small groups to try and find the answer. If any groups are successful too quickly (!) ask them to change the total number of counters, or one or more of the fractions, and to adapt their strategies to the new situations.

As a group discuss the methods used.

What worked? What didn't work?

If faced with a similar problem in future, which methods would the class use?

Here are some possibilities:

- trial and error, making strong use of the upper bound of 40,
- working backwards from the fact that they end up with the same amount.
- focusing on one individual child's initial share of counters
- use the fact that each child passes a certain fraction to their neighbour.
- algebraic representation

### Key questions

Can Ben start with $10$ counters?

### Possible support

Group students in $3$s and provide them with sets of $40$ counters. Ask them to imagine playing the game and challenge them to solve the problem, using these rules:

- Decide who will be Emma, Jack and Ben.
- Ben chooses
*any number of counters*from the $40$ counters and notes this number down - Jack then chooses
*any number of counters from those left over*and notes this number down - Emma then chooses a
*ny number of counters from those left over*and notes this number down - Ben, Jack and Emma then find a third, quarter and fifth of their counters respectively
- All pass the counters to their neighbour: Ben to Jack, Jack to Emma, and Emma to Ben
- If they all end with the same number of counters, they have solved the problem

### Possible extension

Change the numbers. What if the $40$ were $100$, or a limitless supply of counters?