# The Pay Phone Problem

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Introduction

The Pay Phone Problem

Introduction

This coursework is about finding all the possible combinations for putting money in to payphones using various different coins and using those results to try to find a formula that works so you would successfully be able to predict how many coins you would have to put in the payphone for the next total without having to go through all the listings.

Objective

- A woman is going to make a phone call costing any multiple of 10p. I am going to investigate the number of different ways she could put the 10p and 20p coins into the payphone.

2) A man also wants to use the pay phone. He has plenty of 10p and 50p coins. He has no other coins. He wishes to make a telephone call costing any multiple of 10p.

I am going to investigate the number of different ways he has of entering the 10p and 50p coins into the telephone.

3) I will then investigate the more general cases

10p’s and 20p’s

## Results

## Amount Spent (p) | Ways of putting the coins in (p) | Total number of ways |

10 | 10. | 1 |

20 | 20, 10+10. | 2 |

30 | 20+10, 10+20, 10+10+10. | 3 |

40 | 20+20, 20+10+10, 10+20+10, 10+10+20, 10+10+10+10. | 5 |

50 | 20+20+10, 20+10+20, 10+20+20, 20+10+10+10, 10+20+10+10, 10+10+20+10, 10+10+10+20, 10+10+10+10+10. | 8 |

60 | 20+20+20, 20+20+10+10, 20+10+20+10, 20+10+10+20, 10+20+20+10, 10+20+10+20, 10+10+20+20, 20+10+10+10+10, 10+20+10+10+10, 10+10+20+10+10, 10+10+10+20+10, 10+10+10+10+20, 10+10+10+10+10+10+10. | 13 |

## Amount Spent | ## No. of Ways |

10 | 1 |

20 | 2 |

30 | 3 |

40 | 5 |

50 | 8 |

60 | 13 |

Pattern

Middle

2

60

50+10

10+50

10+10+10+10+10+10

3

70

50+10+10

10+50+10

10+10+50

10+10+10+10+10+10+10

4

80

50+10+10+10

10+50+10+10

10+10+50+10

10+10+10+50

10+10+10+10+10+10+10+10

5

90

50+10+10+10+10

10+50+10+10+10

10+10+50+10+10

10+10+10+50+10

10+10+10+10+50

10+10+10+10+10+10+10+10+10

6

100

50+50

50+10+10+10+10+10

10+50+10+10+10+10

10+10+50+10+10+10

10+10+10+50+10+10

10+10+10+10+50+10

10+10+10+10+10+50

10+10+10+10+10+10+10+10+10+10

8

Amount spent (p) | Total no. of ways of putting in coins |

10 | 1 |

20 | 1 |

30 | 1 |

40 | 1 |

50 | 2 |

60 | 3 |

70 | 4 |

80 | 5 |

90 | 6 |

100 | 8 |

## Pattern

The numbers in the table showing the number of ways possible to pay a call costing multiples of ten, using 10p’s and 50p pieces. Using the number of ways:1, 1, 1, 1, 2, 3, 4, 5, 6, 8; I noticed that adding the previous term and the term 5 terms ago gives you the next number in the sequence.

Un=Un-1+Un-5, where Un is the number of ways to make a call costing 10n pence.

## Prediction

Knowing that the previous term and the tem 5 terms ago makes the next one I can predict that for a 110p call the would be 11 different ways of putting the coins in.

## Explanation for Prediction

As every total no of combinations of putting the coins into the pay phone are found by adding the previous term and the term 5 terms ago, the result of Un can be solved by this formula:

Un=Un-1+Un-5, where Un is the number of ways to make a call costing 10n pence.

e.g. for 110p we know it is the equivalent of U11. Let us put this into the formula, to find the number of ways.

U11=U11-1+U11-5

=U10+U6

=8+3

=11

Test

Amount spent (p) |

Conclusion

(p)

Total no. of ways of putting in coins

10

10

1

20

10+10

1

30

10+10+10

1

40

40

10+10+10+10

2

50

40+10

10+40

10+10+10+10+10

3

60

40+10+10

10+40+10

10+10+40

10+10+10+10+10+10

4

So now I have worked out the combinations for; (10p, 20p), (10p, 30p), (10p, 40p) and (10p, 50p) and found the formulae, which are shown below:

Coins for combinations Formulae

10p’s and 20p’s, 10p’s and 30p’s, 10p’s and 40p’s, 10p’s and 50p’s.

Un=Un-1+Un-2, Un=Un-1+Un-3, Un=Un-1+Un-4, Un=Un-1+Un-5.

From the above it is clear that there is a relationship between the coins for combinations and the relevant formulae. The last numbers of the formulae changes to 2, 3, 4 and 5 when we use 20p, 30p, 40p and 50p respectively. This also shows us that when 20 is divided by 10 it equals 2 and 30 divided by 10 equals 3 and so on.

As the 10p is constant throughout the tables, the first part of the formulae (Un-1) is also remaining constant and 10 divided by 10 equals 1.

From these formulae, it is possible to find the General Rule for (10p, np) where; f = any multiple of 10 and fp >10p

So General Rule for (10p, np) would be;

Un=Un-1+Un-f/10

Test

In the following example, I am going to test the total no of combinations for 50p made by 10p and 20p coins. U5 = 50p

(10p, fp)Un = Un-1+Un-f/10

(10p, 20p)U5 = U5-1+U5-20/10

= U4+U3

= 5+3

= 8

This student written piece of work is one of many that can be found in our GCSE Pay Phone Problem section.

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