# Fair Shares?

## Problem

*Fair Shares printable worksheet*

Last weekend Mrs Jenkins won £25 and she gave her winnings to her five children.

She gave her first child **£1** plus $\frac{1}{6}$ of the money remaining.

She gave her second child **£2** plus $\frac{1}{6}$ of the money remaining.

She gave her third child **£3** plus $\frac{1}{6}$ of the money remaining, and so on...**Without doing any calculations, which child do you think ended up with the most money?****Work out how much each child received. Are you surprised?**

Mrs Hobson also had some money to share with her family.

She gave her first child **£1** plus $\frac{1}{5}$ of the money remaining.

She gave her second child **£2** plus $\frac{1}{5}$ of the money remaining.

She gave her third child **£3** plus $\frac{1}{5}$ of the money remaining, and so on...**How much money did she have to share out if all the children received the same amount?****How many children were there in the family?**

In a family with $8$ children, the mother wants to give each child a lump sum plus a fraction of the remainder, in the same way that Mrs Jenkins and Mrs Hobson did.**How much money will she share out, and what fraction will she use each time, in order to share the money equally?****Extension**

Imagine someone had $n$ children.

How much money do you think you would need, in order to give every child an equal share using the method above?

Can you show that the first child will receive $\frac{1}{n}$ of the total?

Can you show that the second child will also receive $\frac{1}{n}$ of the total?

What about the third? The fourth? The $k^{th}$ child? ...

## Getting Started

## Student Solutions

We received a variety of solutions to each part of this problem. Patrick from Woodbridge School explained how he was surprised to find each child received the same amount, and went on to use a spreadsheet to investigate:

At first it seemed that the 5th child would obviously get most, but then I realised that in fact $\frac{1}{6}$ of the money is quite a lot. I then thought that they would all be very closely grouped with child 5 having slightly more. It was surprising to see that every child received £5.

I then adapted the spreadsheet for a varying number of children. I decided that since the denominator of the fraction was decreased by one, I would decrease the number of children by one, so I performed trial and error testing on sums of money equalling 1 mod 5 (I was assuming a simple case of the first child receiving a whole amount of money.) I got a good result with £16 shared between 4 children.

This seems to show a relationship: $\frac{1}{6}$ gives 5 children £5, $\frac{1}{5}$ gives 4 children £4. I decided to test this on the spreadsheet, and the pattern seems to hold that a fraction of $\frac{1}{n}$ gives $n-1$ children £$n-1$ each.

8 children means that $n = 9$ for my calculations, so $\frac{1}{9}$ must be used to share out $8 \times 8 = $ £$64$. I checked this and it worked.

Let there be $n$ children. We must show that $\frac{1}{n+1}$ is the fraction and the total sum of money is $n^2$.

First, to prove that child 1 gets £$n$: The fractional part of his sum is $\frac{n^2-1}{n+1}$, which reduces (because we have a difference of two squares as the numerator) to $n-1$.

Child 1 gets the fractional part plus £1, so his sum is $n-1+1 = n$.

In the case of child $a$, we know that $(a-1) \times n$ has been given out already, so the money left is $n^2-n(a-1)$. Therefore his sum is $\frac{n^2-n(a-1)-a}{n+1} + a$.

This simplifies to $\frac{n^2-na+n-a+an+a}{n+1}$, simplifying to $\frac{n^2+n}{n+1}$ and thence to $n$. Thus, every child gets £$n$ out of a prize fund of $n^2$, if the fraction used is $\frac{1}{n+1}$.

Joshua, from St John's Junior School, explained how you can always find an amount to share in this way with $n$ children:

This only works if you've got $n$ children, £$n^2$ and each time divide by $n+1$.

Child 1 receives: $$1 + \frac{n^2 -1}{n+1} = 1+ \frac{(n-1)(n+1)}{n+1} = 1 + (n-1) = n$$

Child 2 receives: $$2 + \frac{n^2 -n -2}{n+1} = \frac{2(n+1) + (n^2 - n -2)}{n+1} = \frac{n^2 +n}{n+1} = n$$

In general, child $a$ receives: $$a + \frac{n^2 - n(a-1) -a}{n+1} = \frac{n^2 - na + na + a - a + n}{n+1} = \frac{n^2+n}{n+1} = n$$

This means every child gets £n.

Harrison, from Caringbah High School, Australia, and Francois, from Abingdon School, both used simultaneous equations to find out how much Mrs Hobson had to share out. You can read Francois' solution here. Preveina from Crest Girls' Academy sent in this solution. Well done to all of you.

## Teachers' Resources

*This problem is available as a printable worksheet: Fair Shares*

Why do this problem?

This problem starts with a simple situation which can be analysed quickly using mental methods, but which provides a starting point for tackling a more challenging problem. The challenge can be tackled at many different levels, using trial and improvement (perhaps using spreadsheets), looking for number patterns, or with
a more formal algebraic approach.

### Possible approach

Introduce the first problem. Solve it together with the class. Can anyone explain why everything works out so neatly?

Now share the second problem. Students may wish to use trial and improvement to solve it (using spreadsheets might make things quicker). Alternatively, they may use insights from the first problem to suggest starting points. For those who are confident using algebra, the problem can be expressed and solved using equations.

Share approaches and answers to the second problem and encourage the students to use insights from this discussion to help them solve the third problem.

After solving the third problem, students who are less confident with algebra could be encouraged to look at the patterns in the results so far and to suggest generalisations based on what they have seen. Then they could test their conjectures with a variety of examples of their choice.

### Key questions

How can we express each part of this problem algebraically?

### Possible Support

Peaches Today, Peaches Tomorrow.... might offer a suitable introduction to the fraction skills required in this problem.

### Possible Extension

Students who are more confident can be challenged to write their generalisation algebraically, and manipulate the algebraic expressions to produce a rigorous proof that the results will always hold.