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Stage: 3 Challenge Level: Challenge Level:1

Good solutions were received from Luke from St George's College, Ruth from Manchester High School for Girls, Daniel from Junction City High School, Max from Bishop Ramsey C of E Secondary School and Rohan from Long Bay Primary.

Luke wrote:

Since two thirds of the adult men are married, the total number of men must be a multiple of 3.
Possible totals are:

Total Men Total Married Men
3 2
6 4
9 6
12 8

Since three quarters of the adult women are married, the total number of women must be a multiple of 4.
Possible totals are:

Total Women Total Married Women
4 3
8 6
12 9
16 12

The smallest number of couples is 6, when there are 9 men and 8 women.
There would be 17 adults in the smallest community of this type.

Ruth wrote:

There are $m$ men and $w$ women in the community. $\frac{2}{3}$ of the men and $\frac{3}{4}$ of the women are married and there must be the same number of married men as married women, so \begin{eqnarray} \frac{2}{3}m &=& \frac{3}{4}w \\ 8m &=& 9w \end{eqnarray} Now $m$ and $w$ are both integers so the smallest solution to this equation is $$m = 9 , w = 8 $$so there are 17 people in the community, and 6 married couples.

There are other solutions gained by multiplying all the numbers by a constant for example you can double them all to get 18 men, 16 women and 12 married couples but 9 men and 8 women is the smallest solution.

Max wrote:

First of all, I needed the amount of men and women to be married to be equal. Therefore, I needed to find the lowest common numerator.

Taking $\frac{2}{3}$ and $\frac{3}{4}$ and converting them into $\frac{6}{9}$ and $\frac{6}{8}$ shows us that the smallest number of married men and women must be 6.

By adding both of the denominators together, I found that the lowest number of people needed in the community must be 17.