2004 AMC 10A Problem 6
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 grand-daughters have no daughters?
Show solution without answering
Key insight
The 24 granddaughters come in groups of 6, so only 4 daughters are mothers. Everyone else, 30 - 4 = 26 women, has no daughters.
Solution
Sort the women by generation and count each group.
1. Bertha has daughters. The total is , so the number of granddaughters is .
2. Each daughter who has children has exactly of them. So the number of daughters with children is .
3. That leaves daughters with no children.
4. There are no great-granddaughters, so none of the granddaughters has a daughter.
5. The women with no daughters are the childless daughters and all granddaughters:
The answer is .
Why this works
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."
Alternative approach
Count the mothers and subtract. The only women in the group who have daughters are the daughters of Bertha found in step 2. So women have none.
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
Of the women counted, are Bertha's daughters, so are granddaughters.
Each daughter who has children has exactly of them, so the number of daughters with children is . The remaining daughters have no children.
No granddaughter has children (there are no great-granddaughters). So the women with no daughters are the childless daughters together with all granddaughters:
The answer is .
Review in the AoPS books
- Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
- Introduction to Algebra, Chapter 7: Proportion (§7.4)
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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site. Also appeared as 2004 12A #4.