This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### Sum Equals Product

If \(a+b=a\times b\), then:

$$ab-b=a$$
$$b(a-1)=a$$
$$b=\frac{a}{a-1}$$
This will work for any \(a \not= 1\) (\(a=1\) will not work as this will mean division by zero).

##### Extension

(i) Given a number \(a\), can you find a number \(b\) such that \(b-a =\frac{b}{a}\)?

(ii) Given a number \(a\), can you find a number \(b\) such that \(b-a =\frac{a}{b}\)?

(iii) Given a number \(a\), can you find a number \(b\) such that \(a-b =\frac{b}{a}\)?

(iv) Given a number \(a\), can you find a number \(b\) such that \(a-b =\frac{a}{b}\)?

#### Wool Circles

Starting with the person who starts with the wool and going anti-clockwise, number the people \(0,1,2,3,4,...\). As the wool is passed, it will be held by people with numbers:

$$0,a+1,2(a+1),3(a+1),...,k(a+1),...$$
For example, if \(n=10\) and \(a=3\):

The first person will have the wool again when:

$$k(a+1)\equiv 0 \mod n$$
or

$$k(a+1)=ln$$
This will first occur when (hcf is highest common factor):

$$l=\frac{a+1}{\mbox{hcf}(a+1,n)} \mbox{ and } k=\frac{n}{\mbox{hcf}(a+1,n)}$$
\(k\) is also the number of people who are holding the wool. So the number of different coloured balls needed is:

$$\frac{n}{\left(\frac{n}{\mbox{hcf}(a+1,n)}\right)}$$$$= \mbox{hcf}(a+1,n)$$
##### Extension

The ball is passed around the circle of \(n\) people again. This time, the number of people missed alternates between \(a\) and \(b). How many different coloured balls of wool are now needed?

$$y=(1-\sqrt{2x})^2$$