Gcse Maths Coursework: Payphone Problem

40 p

50 p

70 p

80 p

90 p

00 p

Combinations

2

3

5

8

3

21

34

55

89

First Difference

+1

+1

+2

+3

+5

+8

+13

+21

+34

Second Difference

+0

+1

+1

+2

+3

+5

+8

+13

There is no pattern of difference between the different coin amounts so I must have to work out a formula using the adding method (previous page) rather than using differences.

I know that I add the two previous combination amounts to find the number of combinations I want.

nth call cost's combinations = no. of combinations of the call cost before the nth + no. of combinations of the call cost two in front of the nth

If I put this in to algebra, I can put the previous combinations into brackets and then call that bracked 'th' to show that it is the 'x th' term's combinations, not just the 'x th' term.

Also, really, the below is true. Each call cost is just the nth minus a number, so I can use this rule for my formula.

. 1

2. 1

3. 1

4. 1

5. 2 [n - 5 th term]

6. 3 [n - 4 th term]

7. 4 [n - 3 th term]

8. 5 [n - 2 th term]

9. 6 [n - 1 th term]

0.8 [nth term]

nth = (n-1) th + (n-2) th

There is my algebraic formula for part one.

Part 2. Another man has many 10 p and 50 p coins. Investigate all calls in multiples of 10 p.

0 p = 1

20 p = 11

30 p = 111

40 p = 1111

50 p = 11111

5

60 p = 15

51

111111

70 p = 511

151

115

1111111

80 p = 5111

1511

1151

1115

11111111

90 p = 51111

15111

11511

11151

11115

111111111

00 p = 511111

151111

115111

111511

111151

Call Cost

0 p

20 p

30 p

40 p

50 p

60 p

70 p

80 p

90 p

00 p

Combinations

2

3

4

5

6

8

First Difference

0

0

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

2. 1

3. 1

4. 1

5. 2

6. 3

7. 4

8. 5

9. 6

0. 8 Unknown (for the test)

6 + 2 = 8 ---- Correct!

n

. 1

2. 1

3. 1

4. 1

5. 2

6. 3

7. 4

8. 5 Unknown (for test)

9. 6

0. 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.

nth = (n-1) th + (n-5) th

Test:

If n = 10

(10-1) th + (10-5) th

= 8 ---- Correct!

If n = 9

(9-1) th + (9-5) th

= 6 ---Correct!

Part 3. Investigate the more general case (Investigate two more coins --> 3 more coins)

I will investigate 10 p and 100 p (£) coins.

0 p = 1

20 p = 11

30 p = 111

40 p = 1111

50 p = 11111

60 p = 111111

70 p = 1111111

80 p = 11111111

90 p = 111111111

00 p (£) = 111111111

£

Call Cost

0 p

20 p

30 p

40 p

50 p

60 p

70 p

80 p

90 p

00 (£) p

Combinations

Not enough

spaces to count

back 10

2

I will now see if the solution used previously works for this. I counted back five spaces for 50 p coins. So if I count back 10 spaces for this example it should work. However, there aren't enough spaces to go back ten, so I will need to do larger call costs.

10 p = 11111111111

£1

1£

20 p = 111111111111

£11

1£1

11£

30 p = 1111111111111

£111

1£11

11£1

111£

40 p = 11111111111111

£1111

1£111

11£11

111£1

1111£

50 p = 111111111111111

£11111

1£1111

11£111

111£11

1111£1

11111£

Call Cost

0 p

20 p

30 p

40 p

50 p

60 p

70 p

80 p

90 p

00 p

10 p

20 p

30 p

40 p

50 p

Combinations

2

3

4

5

6

7

Now I can convert the previous formula by changing the second value to ten instead of five.

.

nth = (n-1) th + (n-10) th

Test:

n

. 1

2. 1

3. 1

4. 1

5. 1

6. 1

7. 1

8. 1

9. 1

0. 2

1. 3

2. 4

2. 5

4. 6

5. 7

If n = 11

(11-1) 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.

0 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

0 p

20 p

30 p

40 p

50 p

60 p

70 p

Combinations

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

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:

0 p 20 p coins

20 p 50 p coins

0 p 50 p coins

0 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