2004 AMC 12A Problem 4

Attempt Problem 4 of the 2004 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

4.

Bertha has 66 daughters and no sons. Some of her daughters have 66 daughters, and the rest have none. Bertha has a total of 3030 daughters and granddaughters, and no great-granddaughters. How many of Bertha’s daughters and granddaughters have no daughters?
2222
2323
2424
2525
2626
Answer: E
Concepts:basic counting
Small hint:
First count the granddaughters, then determine how many daughters have children
Big hint:
Every woman with daughters has exactly 66 of them
Solution:
Bertha has 30−6=2430 - 6 = 24 granddaughters, none of whom have daughters.
These granddaughters are the children of 246=4\frac{24}{6} = 4 of Bertha’s daughters, so exactly 44 women have daughters.
Therefore the number of women with no daughters is 30−4=26.30 - 4 = 26.
Thus, the correct answer is E.
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