I recently found a card on which there were two puzzles, both of which have me stumped; I want to spend a little more time on the other puzzle, but here is one that stumps me:

no 65
A census-taker calls at a house. He asks the woman living there the ages of her three daughters. The woman says, “If you multiply their ages the total is 72; if you add together their ages the total is the same as the number on my front door, which you can see.”
The census-taker says, “That is not enough information for me to calculate their ages.”
The woman says, “Well, my eldest daughter has a cat with a wooden leg.”
The census-taker replies, “Ah! Now I know their ages.”

What are the ages of the three girls?


Let $d_0,d_1,d_2$ be the ages of the three daughters, with $d_i\leq d_{i+1}$. We know that $d_0d_1d_2=72$ and that $d_0+d_1+d_2=n$, $n$ being the house number. The problem does not give the house number; that is where I am stumped. (HINT: The second-to-last line gives that there is an eldest daughter, making true $d_0\leq d_1<d_2$.)


I think the answer is:

3, 3, and 8. You know the census taker knows the number on the front door and needs to know that there is an eldest daughter to know for sure the correct solution. Therefore, there has to be two combinations of three factors of 72 that have the same sum. Breaking them up, there are only two that sum to the same number:

2+6+6 = 14, 3+3+8 = 14

This means that the first case is eliminated if there is an eldest daughter, and therefore the correct solution is 3, 3, and 8.

| improve this answer | |
  • $\begingroup$ Ahh, I see. Since he needed more information, there must have been more than one solution before-hand. Thanks! $\endgroup$ – Conor O'Brien Jul 12 '15 at 19:14

Not the answer you're looking for? Browse other questions tagged or ask your own question.