2004 AMC 10A Problem 6

Attempt Problem 6 of the 2004 AMC 10A 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 10A solutions, or check the answer key.

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

6.

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 countinglogical deduction
Small hint:
Bertha has 30−6=2430 - 6 = 24 granddaughters, and none of them have daughters
Big hint:
The granddaughters come in groups of 6,6, so find how many daughters are mothers
Solution:
Bertha has 30−6=2430 - 6 = 24 granddaughters, none of whom have daughters.
These granddaughters belong to 246=4\frac{24}{6} = 4 of Bertha’s daughters. So exactly 44 women have daughters, and the number with no daughters is 30−4=26. 30 - 4 = 26.
Thus, the correct answer is E.
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