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Two math teachers were having a chat:

  • Do you have children?
  • Yes, I have 3 sons.
  • What are their ages?
  • Let's make this more fun. If you multiply the ages of my sons, you get the number 36.
  • Ok.
  • And the sum of my sons' ages are.. Hmm.. Do you see the windows of that building?
  • Yes.
  • It's half of the number of those windows.
  • Ok... Hmm... Can you give me one more hint?
  • Sure. My older son's hair is red.

If you were having this chat, what would you say now?

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  • $\begingroup$ It's not quite the same question. It has a distinct answer and a somewhat distinct reasoning. $\endgroup$ – Ben Barden Aug 29 at 17:35
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It's 9/2/2. First, we know that there is an eldest (and therefore not a tie for eldest). We know that we have (multiplicatively) two 2s and two 3s to assign. We know their sum. We can assume (though it's not technically necessarily true) that we're dealing with whole numbers, because otherwise some clues would be meaningless

So, let's look:

(eldest is 1/2/3: no viable sets.) (eldest is 4: only set is 4/3/3: 10 sum.) (eldest is 6: only set is 6/3/2: 11 sum) (eldest is 9: could be 9/2/2 (13) or 9/4/1 (14)) (eldest is 12: only set is 12/3/1 (16)) (eldest is 18: only set is 18/2/1 (21)) (eldest is 36: only set is 36/1/1 (38))

from that

in each case the sum is unique, and thus the ages can be determined, given the sum. But we have another piece of information. The ages could not be determined from sum and multiple alone, which means it would have to have the same sum as a case where there was no "eldest". The only case like that is 6/6/1, which comes to 13 - thus opposing 9/2/2.

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