Congratulations to the many people who sent in excellent solutions. Jeni and Lucy , St James Middle School, Bury St Edmunds and Jackie of Madras College, St Andrews used trial and improvement. All the following used algebra: Timothy of Munsang College, Hong Kong; Kate and Katherine of Emmbrook School, Wokingham; Lyndsey, Christiane and Joanne ofThe Mount School, York; Fiona and Elizabeth of Stamford High School, Lincs.

Here is the solution from Prav, Sheli, Meg, Ruoyi & Liz 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

$\frac{£ 8100}{£ 900} = 9$

there are nine children.

Love from, Prav, Sheli, Meg, Ruoyi & Liz