# Pennies and Tuppences - Isaac's Solution

Thank you to Isaac for sending in this solution to Pennies and Tuppences.

Ways(m) = (n + (m – n (mod m)) + m)/m

This works for coins of 1 pence and m pence, where n is the amount of money for which you are trying to find the amount of money.

This works because for any way to make pence, you can add a 1 to it and get a way to make x+1 pence and the amount of ways to make x doesn’t increase at all, but if x is a multiple of m, a possible way to make x is simply m, 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, which is equal to the next multiple of m divided by m, because everything up to the first multiple of mcan 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 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 mis added to this result, which takes it up to the next multiple of m. Finally, the whole result is divided by to create the amount of ways.
Response to: