We have two responses for you

Hi Paul,

I can get you started on these problems.

There must be a bag with 1 penny in it as you need to be able to pay 1p. To pay 2p you either need another bag with 1 penny in it or a bag with 2 pennies in it. An advantage of having the second bag with 2 pennies in it is that with these two bags you can pay three amounts, 1p, 2p and 3p using both bags.

Can you continue?
Penny

Paul,

This question, and the other one you sent in, are about counting in binary (base two).
With n binary digits it is possible to count to 2^{n+1} - 1. For example, with three binary
digits, it is possible to count to 2^3-1 (seven). Here is the correspondence:

000 = 0x2^2 + 0x2 + 0 = 0
001 = 0x2^2 + 0x2 + 1 = 1
010 = 0x2^2 + 1x2 + 0 = 2
011 = 0x2^2 + 1x2 + 1 = 3
100 = 1x2^2 + 0x2 + 0 = 4
101 = 1x2^2 + 0x2 + 1 = 5
110 = 1x2^2 + 1x2 + 0 = 6
111 = 1x2^2 + 1x2 + 1 = 7

So, for the pennies problem, if there are three bags that contain 1, 2, and 4 pennies,
respectively, then it is possible to pay any amount from 1p to 7p.

I hope this helps you solve both of your problems, and all others like them.

Victoria