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This is what they did:
First of all, we found 15 play coins and a painting tray to help us think about the investigation and find a solution to the maths problem. After that,we decided that we definitely needed one bag with 1 pence in it.or we would not be able to pay for something that cost pence!
Afterwards, we figured out the all the numbers we would need.Wso we put 2p in our second pot. We tried 3p in the third pot and this gave us more prices:
1p (1p); 2p (2p); 3p (3p); 4p (3p + 1p); 5p (3p + 2p); 6p (3p + 2p + 1p)
...but we couldn’t make 7.
So we tried again, this time with 4p in the third pot. We could then make all of the prices up to 7p.
1p (1p); 2p (2p); 3p (2p + 1p); 4p (4p); 5p (4p + 1p); 6p (4p + 2p); 7p (4p + 2p + 1p).
1p (1p); 2p (2p); 3p (2p + 1p); 4p (4p); 5p (4p + 1p); 6p (4p + 2p); 7p (4p + 2p + 1p).
So we tried again, this time with 4p in the third pot. We could then make all of the prices up to 7p.
1p (1p); 2p (2p); 3p (2p + 1p); 4p (4p); 5p (4p + 1p); 6p (4p + 2p); 7p (4p + 2p + 1p).
So we knew we had found the solution! We tried all the possibilities and didn’t give up because Cabot Primary children are resilient! PS Our paint tray actually has six pots in it. We can see a pattern building up and are now going to think about how to investigate the maximum number of prices we can make if we use all six pots.
When we looked again at the number in each pot, we noticed that each number was double the previous one. Since double 8 is 16, we put 16 coins into the next empty pot so we had five pots containing, 1p,2p,4p,8p,and 16p. We knew the maximum number we could now pay is 31p (our existing 15p plus the extra pot of 16p). We checked that we could now make all of the prices from 15p to 31p (which we could – by adding the new pot of money to each of the combinations that we had already found). We knew that we had found a system and that the sixth pot should therefore contain 32p (which is double 16p) and that we would be able to make all of the prices up to a maximum of 63p.We then came up with this sequence to show how much money twelve pots (two paint trays) would contain:
1p, 2p, 4p, 8p, 16p, 32p, 64p, £1.28, £2.56, £5.12, £10.24 £20.48!
1p (1p); 2p (2p); 3p (2p + 1p); 4p (4p); 5p (4p + 1p); 6p (4p + 2p); 7p (4p + 2p + 1p).
When we looked again at the number in each pot, we noticed that each number was double the previous one. Since double 8 is 16, we put 16 coins into the next empty pot so we had five pots containing, 1p,2p,4p,8p,and 16p. We knew the maximum number we could now pay is 31p (our existing 15p plus the extra pot of 16p). We checked that we could now make all of the prices from 15p to 31p (which we could – by adding the new pot of money to each of the combinations that we had already found). We knew that we had found a system and that the sixth pot should therefore contain 32p (which is double 16p) and that we would be able to make all of the prices up to a maximum of 63p.We then came up with this sequence to show how much money twelve pots (two paint trays) would contain:
1p, 2p, 4p, 8p, 16p, 32p, 64p, £1.28, £2.56, £5.12, £10.24 £20.48!
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