Puzzles

The digits of the number must be 1, 2 and 3, so the largest number is 321.

The two numbers are 243 and 244.

The smallest number you can make with the digits in the red boxes is 123.

Because 13 and 11 are prime and larger than 7, no-one can have taken them, as their is no way the other person's product could then be the same. Once 13 and 11 are discarded, the prime factorisation of the product of all the other cards is \(2^{11}\times3^5\times5^2\times7^2\). In order to be shared to give the same product, the powers must be even, and so the other card not taken must be 6 (as 2 and 3 are the numbers with odd powers).

Therefore the product of the three cards is 858.

The diagonal of the blue square is \(\sqrt{28}\). The radius \(R\) of the large circle satisfies \(R^2+R^2=(R+\sqrt{28})^2\). Solving this, we find that \(R=\frac{\sqrt{28}}{\sqrt2-1}=\sqrt{28}(\sqrt2+1)\).

The radius \(r\) of the small circles satisfies \(r+r\sqrt2=R\), and so \(r=\frac{\sqrt{28}(\sqrt2+1)}{\sqrt2+1}=\sqrt{28}\).

The area of the square is \(4r^2=4\left(\sqrt{28}\right)^2=4\times28\). This is 112.

To win this game, you need to always end your turn on a multiple of 901, as whatever you opponent does, you can always get back to a multiple of 901.

Therefore, you should take £791 on your first turn to get to a multiple of 901.

For the first and last digits of the result to be different, there must be a 1 carried from the middle column. Therefore the middle digit of Eve's number is 9. Eve's number could be 292 or 193, but it cannot be the first as reversing this is not larger than her original number, so here number is 193.