2004 AMC 10A Problem 6

Sort the women by generation and count each group.

1. Bertha has $6$ daughters. The total is $30$, so the number of granddaughters is $30 - 6 = 24$.
2. Each daughter who has children has exactly $6$ of them. So the number of daughters with children is $24 \div 6 = 4$.
3. That leaves $6 - 4 = 2$ daughters with no children.
4. There are no great-granddaughters, so none of the $24$ granddaughters has a daughter.
5. The women with no daughters are the $2$ childless daughters and all $24$ granddaughters:
$$2 + 24 = 26.$$

The answer is $\boxed{\textbf{(E)}\ 26}$.

Splitting a family into generations turns a confusing sentence into simple counts. Because every mother has the same number of daughters, one division tells you how many mothers there are. The rest is just "everyone minus the mothers."

Count the mothers and subtract. The only women in the group who have daughters are the $4$ daughters of Bertha found in step 2. So $30 - 4 = 26$ women have none.

Of the $30$ women counted, $6$ are Bertha's daughters, so $30 - 6 = 24$ are granddaughters.

Each daughter who has children has exactly $6$ of them, so the number of daughters with children is $24 / 6 = 4$. The remaining $6 - 4 = 2$ daughters have no children.

No granddaughter has children (there are no great-granddaughters). So the women with no daughters are the $2$ childless daughters together with all $24$ granddaughters:
$$2 + 24 = 26.$$

The answer is $\boxed{\textbf{(E)}\ 26}$.