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?

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