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Comment by dctoedt | Hacker News Reader

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dctoedt

> 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.)

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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.

> 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.

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.

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