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Gcse Maths Coursework: Payphone Problem

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Introduction

GCSE Maths Coursework: Payphone Problem A payphone only accepts 10p, 20p, 50 p and �1 coins. 1 stands for 10p 2 stands for 20 p 5 stands for 50p � stands for one pound In order to make sure I work out all the combinations correctly I will use a logical system. I will start with the call cost made up of the smallest coin type then add in the larger coin at the start and move it across one position for each combination. I will repeat this, changing the coin orders until it is needed Part 1. A woman has plenty of 10 p and 20p coins. She can put them in the phone in any order. Investigate the ways of making any call in multiples of 10 p. 10 p = 1 20 p = 11 2 30p = 111 21 12 40 p = 1111 211 121 112 22 50 p = 11111 2111 1211 1121 1112 221 122 212 Call Cost 10 p 20 p 30 p 40 p 50 p Combinations 1 2 3 5 8 Prediction 60 p = 5 + 8 = 13 I predict that 60 p will have 12 combinations because the two previous numbers of combinations added together make the number of combinations after them. Check: 111111 21111 12111 11211 11121 11112 2211 1221 1122 2121 1212 2112 222 It works, I can now work out the combinations for part one. ...read more.

Middle

1 1 1 1 2 3 4 5 6 8 First Difference 10 0 1 1 2 When doing part 1 also found that there wasn't a pattern of differences between the combinations. However, last time adding the number behind the one I want to find, to the number two spaces behind the one I want to find gave me the number of combinations. That was working with 20p coins, so if I'm working with 50 p coins I might have to go back five places. Let's test my theory... n 1. 1 2. 1 3. 1 4. 1 5. 2 6. 3 7. 4 8. 5 9. 6 10. 8 Unknown (for the test) 6 + 2 = 8 ---- Correct! n 1. 1 2. 1 3. 1 4. 1 5. 2 6. 3 7. 4 8. 5 Unknown (for test) 9. 6 10. 8 4 + 1 = 5 ---- Correct ! My theory seems to work but I can't calculate the number of combinations for 50 p, 40 p, 30 p, 20 p and 10 p because I can't count back five. I am not worried about this because those amounts don't have combinations over 2, so it is very easy to calculate them. I will now try and make an algebraic formula for this part... From the last part I know the formula for going back two spaces, so all I need to do is substitute the five for the two, and I have a formula for this part. ...read more.

Conclusion

th + (11-10) th 2 + 1 = 3 ---- Correct! If n = 12 (12-1) th + (12-10) th 3 + 1 = 4 ---- Correct! I will now investigate 10p and 50 p coins. 10 p = 1 20 p = 11 30 p = 111 40 p = 1111 50 p = 11111 5 60 p = 111111 51 15 70 p = 1111111 511 151 115 Call Cost 10 p 20 p 30 p 40 p 50 p 60 p 70 p Combinations 1 1 1 1 2 3 4 Now I can convert the previous formula by changing the second value to five from ten. . nth = (n-1) th + (n-5) th Test: n 1. 1 2. 1 3. 1 4. 1 5. 2 6. 3 7. 4 If n = 6 nth = (6-1) th + (6-5) th 2 + 1 = 3 --- Correct! If n = 7 nth = (7-1) th + (7-5) th 3 + 1 = 4 --- Correct! General Formula By adapting the formula I can solve: 10 p 20 p coins 20 p 50 p coins 10 p 50 p coins 10 p 100p (�) coins Each time I just substituted the right value you need to count up into the formula. So I can make a general formula along this principle. nth = (n-1) th + (n-x) th x = the first number in the larget coin of a two figure coin. Or the first two figures in a three figure coin (100 p = �) Stephen Webb 1 ...read more.

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