A man gets to know his neighbor, and he learns that she has 3 girls. Intrigued, he asks for their age. The lady, who doesn't want to give him this information straight away, has him to take a guess.

Lady : The product of their ages is 36.

Man : I definitely need more clues...

L : The sum is equal to the number of the house across the street.

M : OK, so, you're not helping, uh?

L : The elder one is fair-haired.

M : ... Oh! I know!

How old are the three girls, and how could the man guess it?

Not sure about my english and/or the tags (first post here, hope it is not a duplicate), so feel free to edit.

  • 2
    $\begingroup$ puzzling.stackexchange.com/questions/208/… $\endgroup$ – Carl Löndahl Jul 4 '16 at 12:13
  • $\begingroup$ @Beastly Gerbi: This is not a duplicate. In the other problem, the product is 72. Here the product is 36. Even the solutions of the two problems are different. $\endgroup$ – Gamow Apr 14 '17 at 19:43
  • 1
    $\begingroup$ @Gamow different numbers same puzzle. You solve the same issue, with slightly different information. Therefore it is a duplicate. The only thing that is different here is the surface, the underlying puzzle is the same. I'm sure IAmInPLS didn't know it was a dupe, but that doesn't mean it isn't one. Just because a solution is different doesn't mean the puzzle is $\endgroup$ – Beastly Gerbil Apr 14 '17 at 21:26

They have to be:

9, 2 and 2 years old

Based on the first information, the possible ages are:

36 1 1
18 2 1
12 3 1
9 4 1
9 2 2
6 6 1
6 3 2
4 3 3

The second bit of information suggests that

the sum of ages is not unique. As the sums of the previous ages are 38, 21, 16, 14, 13, 13, 11 and 10 respectively, it means, the ages have to be either 9, 2, 2 or 6, 6, 1.

The last piece of information tells that

there is a unique oldest child, so 6, 6, 1 is ruled out.

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  • $\begingroup$ That's the solution ! $\endgroup$ – IAmInPLS Jul 4 '16 at 12:22
  • $\begingroup$ This looks right - deleted my answer $\endgroup$ – abligh Jul 4 '16 at 12:25
  • $\begingroup$ ...except that the last piece of information does not rule out any case - e.g. $6\frac34\gt6\frac12$. $\endgroup$ – Jonathan Allan Jul 5 '16 at 8:58
  • $\begingroup$ I feel that would be a bit of a stretch. Not only because it is biologically impossible for the lady to have children that close in age to each other - this is much more a math problem than a lateral thinking one. But also because it would be strange to once use the fractional part of the ages, but neglect them when calculating the sums and products. And I have the intuition, that if non-integer ages are also considered valid, there are too many (maybe an infinite number of) valid combinations. $\endgroup$ – elias Jul 5 '16 at 11:36
  • $\begingroup$ In other words: according to the story, the man knew the ages of the children after the conversation. This would be impossible if non-integers are considered valid (for example $6+3\sqrt{2}$, $2$, $6-3\sqrt{2}$ would produce the same product and sum as $9$, $4$, $1$). So they aren't. $\endgroup$ – elias Jul 5 '16 at 11:51

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