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. Let us put this into the formula, to find the number of ways.
U7=U7-1+U7-2
=U6+U5
=13+8
=21
Test
This shows that my prediction was correct now I will move on to 10p’s and 50p’s.
10p’s and 50p’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 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
This shows that my prediction was correct now I will move on to objective 3. I will invent a 30p coin and find the formula along with a 40p coin. I will then try and find a formula to find the number of ways possible to put the coins into the payphone in general.
10p’s and 30p’s
Pattern
The numbers in the table showing the number of ways possible to pay a call costing multiples of ten, using 10p’s and 30p pieces. Using the number of ways:1, 1, 2, 3, 4, 6, 9; I noticed that adding the previous term and the term 3 terms ago gives you the next number in the sequence.
Un=Un-1+Un-3, where Un is the number of ways to make a call costing 10n pence.
Prediction
Knowing that the previous term and the term 3 terms ago makes the next one I can predict that for a 80p call the would be 13 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 3 terms ago, the result of Un can be solved by this formula:
Un=Un-1+Un-3, where Un is the number of ways to make a call costing 10n pence.
e.g. for 80p we know it is the equivalent of U8. Let us put this into the formula, to find the number of ways.
U8=U8-1+U8-3
=U7+U5
=9+4
=13
Test
This shows that my prediction was correct I will now predict the formula for finding the number of ways to put the coins into the payphone using 10p’s and 40p’s.
Prediction for formula using 10p’s and 40p’s
As we know the formula for 10p’s and 30p’s and 50p’s:
Un=Un-1+Un-3
Un=Un-1+Un-5
I predict that the formula for 10p’s and 40p’s is Un=Un-1+Un-4, where Un is the number of ways to make a call costing 10n pence.
Test
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