Congratulations to the many people who sent in excellent solutions. Jeni Beacock and Lucy Gallagher, (Y7) St James Middle School, Bury St Edmunds and Jackie Matthews (Class 1H) Madras College, St Andrews used trial and improvement. All the following used algebra: Timothy Hung, Munsang College, Hong Kong; Kate Lister and Katherine Taylor of Emmbrook School, Wokingham; Lyndsey Coo, Christiane Eaves and Joanne Barker, (Y10), The Mount School, York; Fiona Watson and Elizabeth Daniels, (Y9) Stamford High School, Lincs.

Here is the solution from Prav, Sheli, Meg, Ruoyi & Liz (Yr. 9) The North London Collegiate School Puzzle Club.

Dear Cambridge,

We have a solution to the Rollerball question published in the May problems. We made a formula to find out the prize money which we called £x.

$\begin{matrix} 100 + \frac{1}{10} (x - 100) & = & 200 + \frac{1}{10} [(x - 300) - \frac{1}{10} (x - 100)] \\ 100 + \frac{x}{10} - 10 & = & 200 + \frac{1}{10} (x - 300 - \frac{x}{10} + 10) \\ \frac{x}{10} + 90 & = & 200 + \frac{1}{10} (- 290 + \frac{9 x}{10}) \\ \frac{x}{10} + 90 & = & 200 - 29 + \frac{9 x}{100} \\ \frac{x}{10} + 90 & = & 171 + \frac{9 x}{100} \\ \frac{x}{10} & = & 81 + \frac{9 x}{100} \\ \frac{x}{100} & = & 81 \\ x & = & 8100 \end{matrix}$

The first child gets:

$100 + \frac{1}{10} (8100 - 100) = 900 .$

Therefore, if every child gets the same amount, so

$\frac{# x 00 A 3; 8100}{# x 00 A 3; 900} = 9$