String Patterns

Introduction Polygons and Stars Cyclic groups Snooker on a circular table Introduction

Using a circular nail board tie a loop in one end of the wool and attach it to one nail. Choose a number and step around the circle, jumping that many nails each time.

For example, suppose there are 10 nails, labelled as 0-9

If we choose to jump 4 nails at a time we would start at 0 and then travel anticlockwise around the circle 0, 4, 8... as shown on the right.

You can see that the sequence 0, 4, 8, 12, 16, 20

corresponds to the nails 0, 4, 8, 2, 6, 0

Can you see that we are just ignoring multiples of 10?

After 5 steps we are back to the start.

We've gone twice around the circle, or 720 degrees.

You can use the activity below to investigate what happens for different numbers of nails and different sizes of jump.

How can we work out what shape we get, and when the wool reaches the 0 nail again?

We need to use the idea of modular arithmetic.

We saw above that 11 = 1 when there are 10 nails.

Mathematically, we write this as 11 = 1 (mod 10) and we say that "11 equals one mod 10".

This is because the remainder when 11 is divided by 10 is 1, or that 11 is one more than a multiple of 10.

Similarly,

• 23 = 3 (mod 10)       23 has remainder 3 mod 10
• 16 = 2 (mod 7)        16 has remainder 2 mod 7
• 40 = 5 (mod 15)        40 has remainder 5 mod 15

We can also say that:

• 60 = 21 (mod 13) 60 and 21 are equal mod 13, since they are both equal to 8 mod 13
• 27 = 81 (mod 9) 27 and 81 are equal mod 9, since they are both equal to 0 mod 9
• A number is a multiple of 'k' when it is equal to 0 (mod k)

If there are 'n' nails and we jump 'a' each time, then our sequence of nails is:

0, a, 2a, 3a, 4a . . . . .

Where each of the numbers is meant to be taken mod n.

If we get back to the start after 'k' jumps we will have a 'k' sided shape (a polygon or a star).

We get back to the start when ka = 0 (mod n).

So, ka is a multiple of n - this means that k needs to contain all the prime factors of n which a doesn't contain.

As an example of this, consider n = 10, and a = 4 from before.

10 has the prime factors 2 and 5, and 4 contains the 2 but not the 5.

So in this case we need k to be 5.

The hcf (highest common factor) of 'n' and 'a' is the product of all common prime factors of n and a.

k must be the product of all the other prime factors of n, so

k =	n   hcf(a,n)

Let's check this works out for that one we looked at before,

n = 10 and a = 4

Our formula predicts that

k =

10     hcf(4,10)

=

10     2

=  5

What about the 720 degrees?

Well we can explain that with our formula too,

ka = some multiple of n           -    Equation 1

The multiple is obviously just the number of times we go round.

Rearranging equation 1, the multiple must be

ka n

But using our formula for 'k', this is equal to

a hcf(a,n)

Let's check when n = 10, a = 4

Our formula says that we go round a total of 4/hcf(10,4) = 2 times.

That's correct - in fact, the sequence was 0, 4, 8, 12, 16, 20

The string joins back on itself at nail 20, which is nothing but nail 0 after going round twice!

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