2004 AMC 10A Problem 6

Below is the professionally curated solution for Problem 6 of the 2004 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10A solutions, or check the answer key.

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Concepts:basic counting logical deduction

Difficulty rating: 1170

Bertha has $6$ daughters and no sons. Some of her daughters have $6$ daughters, and the rest have none. Bertha has a total of $30$ daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and granddaughters have no daughters?

$22$

$23$

$24$

$25$

$26$

Solution:

Bertha has $30 - 6 = 24$ granddaughters, none of whom have daughters.

These granddaughters belong to $24 / 6 = 4$ of Bertha's daughters. So exactly $4$ women have daughters, and the number with no daughters is $30 - 4 = 26.$

Thus, the correct answer is E.

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