2. In the UK, $1 \mathrm{p}, 2 \mathrm{p}$ and $5 \mathrm{p}$ coins have thicknesses of $1.6 \mathrm{~mm}, 2.05 \mathrm{~mm}$ and $1.75 \mathrm{~mm}$ respectively.
Using only 1 p and 5 p coins, Joe builds the shortest (non-empty) stack he can whose height in millimetres is equal to its value in pence. Penny does the same but using only $2 \mathrm{p}$ and 5 p coins.
Whose stack is more valuable?
SolUtion
Say Joe has $a 1 \mathrm{p}$ coins and $b 5 \mathrm{p}$ coins. Then he wants the minimal $a$ and $b$ such that
$$
a+5 b=1.6 a+1.75 b \text {. }
$$

This simplifies to
$$
12 a=65 b .
$$

Since $a$ and $b$ are integers, and 12 and 65 have no factors in common (other than 1 ), $a$ must be a multiple of 65 and $b$ must be a multiple of 12 . Clearly the smallest (positive) $a$ and $b$ which satisfy this equation are $a=65$ and $b=12$. So Joe's stack comprises sixty-five 1 p coins and twelve 5 p coins, giving it a value of 125 p.

Now say Penny has $c 2$ p coins and $d 5$ p coins. Then similarly, she wants the minimal $c$ and $d$ such that
$$
2 c+5 d=2.05 c+1.75 d,
$$
which simplifies to
$$
c=65 d .
$$

The smallest positive $c$ and $d$ which satisfy this equation are $c=65$ and $d=1$. So Penny's stack comprises sixty-five 2 p coins and one 5 p coin, giving it a value of 135 p.
Hence Penny's stack is more valuable.