Question 56

Suppose it is Bec's turn and there are $4$ game pieces remaining. No matter how many pieces Bec chooses, Albert will always be able to pick up the last piece in the next turn.

To guarantee this scenario, there must have been $8$ pieces remaining in Bec's previous turn.

Similarly, there must have been $12$ pieces remaining Bec's turn before that.

So to ensure that he always wins, Albert should make sure that the number of pieces remaining after each of his turns is a multiple of $4$.

Each man bought $m$ sheep at $$m$ each, spending $$m^{2}$. Also each woman bought $w$ sheep at $$w$ each, spending $$w^{2}$.

Each man spent $$63$ more than his wife, so

Since $63$ only has $3$ factor pairs $\quad 63 \times 1$, $\quad 21 \times 3 \quad $and $ \quad 9 \times 7 \quad $, and $m$ and $w$ are positive integers, we can work out $m$ and $w$ by substituting the factor pairs into $\left( m + w\right)$ and $\left( m - w\right)$:

$\therefore\quad $ the men bought $32$, $12$ and, $8$ sheep, and the women bought $31$, $9$, and $1$ sheep.

Since Arthur bought $23$ more sheep than Stef, Arthur bought $32$ sheep, and Stef bought $9$ sheep.

Since Brian bought $11$ more sheep than Rachel, Brian bought $12$ sheep, and Rachel bought $1$ sheep.

Therefore, Charles bought $8$ sheep, and Tracy bought $31$ sheep.

As $\quad 32^{2} - 31^{2} = 63$, $\quad 12^{2} - 9^{2} = 63 \quad $ and $ \quad 8^{2} - 1^{2} = 63$,$\quad$ Arthur is married to Tracy, Brian is married to Stef, and Charles is married to Rachel.