### Counting Factors

Is there an efficient way to work out how many factors a large number has?

### Summing Consecutive Numbers

15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?

### Helen's Conjecture

Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?

# Small Change

##### Age 11 to 14 Challenge Level:

In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins? Remember, the aim is not just to get the answer but to find a good method and to explain it well.

There are more than 4000 possibilities so when you try this question you will find that counting all possibilities is too tedious unless you have a good system to reduce the work and a good notation to record work in progress. If you are going to get the answer you will need to find a good method which you can explain clearly.

Here is one method you might like to follow. Using a spreadsheet saves work but it is still easy to do without one.

Use the notation 100(1,2,5,10,20,50) for the number of combinations of the listed smaller coins which make up one pound and similarly for smaller amounts, for example 30(1,2,5) is the number of combinations of 1p, 2p and 5p coins which make up 30p.

Step 1 Show that the number of ways of changing X pence into 1p and 2p coins is (X/2 + 1) when X is even and (X + 1)/2 when X is odd. Now fill in column A in the table below.

Step 2 Fill in column B in the table using the results in column A and using the earlier results as you work your way down the column. For example we can make up 10 pence using no 5p coins, or one 5p coin or two 5p coins, hence:

10(1,2,5) = 10(1,2) + 5(,1,2) + 1 = 6 + 3 + 1 = 10

and similarly to make up 20 pence we use zero, one, two, three or four 5p coins giving:

20(1,2,5) = 20(1,2) + 15(1,2) + 10(1,2) + 5(1,2) + 1 = 11 + 8 + 6 + 3 + 1 = 29

Step 3 Fill in column C where, for example, corresponding to zero, one, two and three 10p coins we get:

30(1,2,5,10) = 30(1,2,5) + 20(1,2,5) + 10(1,2,5) + 1 = 58 + 29 + 10 + 1 = 98

Step 4 Now you should be able to continue in this manner to fill in the whole table and to get the answer in the bottom right hand corner.

Table showing the numbers of combinations of smaller coins to make up the amounts shown:

A B C D E
5(1,2)=
10(1,2)= 10(1,2,5)=10 10(1,2,5,10)= 10(1,2,5,10,20)= 10(1,2,5,10,20,50)=
15(1,2)=
20(1,2)= 20(1,2,5)=29 20(1,2,5,10)= 20(1,2,5,10,20)= 20(1,2,5,10,20,50)=
25(1,2)=
30(1,2)= 30(1,2,5)=58 30(1,2,5,10)=98 30(1,2,5,10,20)= 30(1,2,5,10,20,50)=
35(1,2)=
40(1,2)= 40(1,2,5)= 40(1,2,5,10)= 40(1,2,5,10,20)= 40(1,2,5,10,20,50)=
45(1,2)=
50(1,2)= 50(1,2,5)= 50(1,2,5,10)= 50(1,2,5,10,20)= 50(1,2,5,10,20,50)=
55(1,2)=
60(1,2)= 60(1,2,5)= 60(1,2,5,10)= 60(1,2,5,10,20)= 60(1,2,5,10,20,50)=
65(1,2)=
70(1,2)= 70(1,2,5)= 70(1,2,5,10)= 70(1,2,5,10,20)= 70(1,2,5,10,20,50)=
75(1,2)=
80(1,2)= 80(1,2,5)= 80(1,2,5,10)= 80(1,2,5,10,20)= 80(1,2,5,10,20,50)=
85(1,2)=
90(1,2)= 90(1,2,5)= 90(1,2,5,10)= 90(1,2,5,10,20)= 90(1,2,5,10,20,50)=
95(1,2)=
100(1,2)= 100(1,2,5)= 100(1,2,5,10)= 100(1,2,5,10,20)= 100(1,2,5,10,20,50)=

There are other ways to do this and you might like to find a different method of your own, perhaps writing a computer program to find the result, do let us know. It would be cool to publish several different methods.