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{matrix} \frac{2}{3} m & = & \frac{3}{4} w \\ 8 m & = & 9 w \end{matrix}

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.