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Comment by cstuder | Hacker News Reader

If you want a problem in similar spirit:

Two mathematicians are meeting again after not having seen each other for a long time. They are discussing their lives, and mathematician A reveals to have three children. Mathematician B ask how old they are.

A: If you multiply their ages, you get 36. And if you sum up their ages, the result is actually the last two digits of your phone number.

She thinks for a moment, but then...

B: That's not enough information, can you tell me more?

A: Yes, you're right. Well, my oldest kid has blue eyes.

How old are A's kids?

(Yes, you can assume their ages to be integers.)

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For this puzzle to work, you can't just assume their ages to be integers - their ages have to be integers. In other words, for this puzzle to be solvable, two people who are both X years old have to be considered to be exactly the same age, with neither elder than the other.

> How old are A's kids?

Much easier than the parent problem (at least now that we've seen how the parent problem unfolds).

Kids' ages: X, Y, and Z.

X * Y * Z = 36

CORRECTED: X + Y + Z ≥ 10 (the sum is two digits) [I had the ≥ backwards].

One triplet that works is 3, 3, and 4; also 1, 4, and 9.

EDIT: OK, I blew this one, I think --- see 'sunlr's correction below. (Note to 'sunlr: The problem as given is that the oldest kid has blue eyes, not the youngest. That fact doesn't eliminate twins --- all the kids could have blue eyes.)

I think the blue eye bit should be the "youngest has blue eyes" and the answer eliminates twins. Thus 2, 3, 6.

Also you have the sign reversed. Should be x + y + z >= 10.

> the answer eliminates twins

I don't believe so, because regardless of which child has blue eyes, it doesn't eliminate the possibility that either or both of the other children also have blue eyes. If the statement was qualified with an "only", then you could eliminate twins.

> X + Y + Z ≤ 10 (the sum is two digits)

I don't understand this part. Surely every number between 10 and 99 (inclusive) has two digits? And numbers <10 have only one digit? Have I forgotten how to count to 2?

If we use your interpretation then the fact there is an oldest child is redundant; but if we don't then the fact that the sum is two digits is redundant, and there are multiple solutions.

The 'oldest child' thing rules out 6, 6, 1 as the answer, but afaict 18, 2, 1 and 12, 3, 1 and 9, 4, 1 and 9, 2, 2 and 6, 3, 2 are still valid solutions as well as 4, 3, 3.

> And numbers <10 have only one digit?

Not in a phone number.

3,3,4 isn't a valid solution because then B would not reply "that's not enough information." There's exactly one triplet that adds up to eleven, so if B's phone number ended in 11, she would have immediately deduced the solution instead of asking for more information. Same with 1,4,9: that's the only triplet adding to 14. The only valid triplets are ones that share a sum with another triplet. Only two triplets multiplying to 36 share a sum: 1,6,6 and 2,2,9 both add up to 13. Once B learns that A has an "oldest kid", B can rule out 1,6,6, because A would have said "one of my oldest kids has blue eyes" in that case. That leaves 2,2,9 as the only valid triplet.

Thank you for this little riddle, I like how it sounds almost like a joke until you realise what the last piece of info actually means.

The ages can be:

    3 + 3 + 4 = 10
    1 + 4 + 9 = 14
    2 + 2 + 9 = 13
    1 + 6 + 6 = 13

If the sum is 10 or 14, B would have had enough information for the answer at once. So the sum should be 13.

And "oldest kid" is not "oldest kids", so 2, 2, 9 is the answer.

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