The Pay Phone Problem

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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

  1. 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

Pattern 

The numbers in the table showing the number of ways possible to pay a call costing multiples of ten, using 10p’s and 20p pieces. Using the number of ways: 1 2 3 5 8 13; I noticed that adding the two previous numbers gives you the next number in the sequence (this is called the Fibonacci sequence).

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Un=Un-1+Un-2, where Un is the number of ways to make a call costing 10n pence.

Prediction

Knowing that the previous two terms make the next one I can predict that for a 70p call the would be 21 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 2 previous terms, the result of Un can be solved by this formula:

Un=Un-1+Un-2

e.g. for 70p we know it is the equivalent of U7. ...

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