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.