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## Monday, 21 July 2014

### Sunday Afternoon Maths XXI Answers & Extensions

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