Charlie would have to carry around 6 coins - either four 2ps and two 1ps or five 2ps and one 1p. The first option would work like this:

1 - 1p

2 - 2p

3 - 2p+1p

4 - 2p+2p

5 - 2p+2p+1p

6 - 2p+2p+2p

7 - 2p+2p+2p+1p

8 - 2p+2p+2p+2p

9 - 2p+2p+2p+2p+1p

10 - 2p+2p+2p+2p+1p+1p

The second option would work like this:

1 - 1p

2 - 2p

3 - 2p+1p

4 - 2p+2p

5 - 2p+2p+1p

6 - 2p+2p+2p

7 - 2p+2p+2p+1p

8 - 2p+2p+2p+2p

9 - 2p+2p+2p+2p+1p

10 - 2p+2p+2p+2p+2p

Alison, however, would only have to carry 5 coins - two 4ps and three 1ps. It would work like this:

1 - 1p

2 - 1p+1p

3 - 1p+1p+1p

4 - 4p

5 - 4p+1p

6 - 4p+1p+1p

7 - 4p+1p+1p+1p

8 - 4p+4p

9 - 4p+4p+1p

10 - 4p+4p+1p+1p

When trying to reach all amounts up to 10p, I would choose either a 3p or a 4p coin because they both only need 5 coins, whereas all other amounts need more than that.

When trying to reach all amounts up to £1, I would choose any number from 8 - 13 because I worked out the formula:

100/n + (n-1)= a

If you do not get a whole number for the division part, then round it down to the next integer.

a= the amount of coins needed.

For the numbers 8 - 13 I needed 19 coins, which is the least amount of coins possible.

Response to: