Pennies and Tuppences

Pile of 2p coins

Charlie won some 1p and 2p coins on the Penny Falls games. When he played around with the coins, he found he could make 8p in five different ways.

Can you find them all? Click below to see:

1p, 1p, 1p, 1p, 1p, 1p, 1p, 1p

1p, 1p, 1p, 1p, 1p, 1p, 2p

1p, 1p, 1p, 1p, 2p, 2p

1p, 1p, 2p, 2p, 2p

2p, 2p, 2p, 2p


Is there a quick way of working out the number of different ways of making any amount with just 1p and 2p coins?

What if we just had 1p and 3p coins instead? Or just 1p and 4p coins? Or ...



Suppose Ways(m,n) describes the number of ways to make a value of n pence using $1$ and $n$ pence coins. Then,
$$\text{Ways}(m,n) = \frac{n + m – n (\mod m)}{m}$$

This works because for any way to make $x$ pence, you can add a $1$ to it and get a way to make $x+1$ pence and the amount of ways to make it doesn’t increase at all. But, if $x$ is a multiple of $m$, a possible way to make $x$ is simply using $m$ pence coins, so the possible ways of making $x$ increases by one.

So all numbers between two multiples of m, including the lower multiple, have the same amount of ways to create them. This is equal to the next multiple of m divided by m, because everything up to the first multiple of m can be made in one way – adding 1s together, then everything from that multiple of m to the next one can be made by 1s, or 1s and an m, giving one extra way. Up to the first multiple of m, there is one way, which is the first multiple of m divided by m, so, as every time you go past a multiple of m, you add another way, the way of making each number is the next multiple of m divided by m.

In my formula, the $m – n (\mod m)$ part creates a number, which when added to $n$, brings it up to the next multiple of m. Finally, the whole result is divided by m to create the amount of ways.