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 aspreadsheetto
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: $$1 + \frac{n^2 -1}{n+1} = 1+
\frac{(n-1)(n+1)}{n+1} = 1 + (n-1) = n$$
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' solutionhere. Preveina from
Crest Girls' Academy sent inthissolution. Well
done to all of you.