What is the Mystery of the Three Sons' Ages?

  • Context: High School 
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SUMMARY

The puzzle of the three sons' ages revolves around the product of their ages being 36 and the sum corresponding to a building number, which remains ambiguous. The two viable combinations that fit these criteria are (2, 2, 9) and (1, 6, 6). The final clue about the oldest son having red hair indicates that there is a distinct oldest child, confirming the ages must be (2, 2, 9) since (1, 6, 6) would imply twins. Thus, the solution is definitively 2, 2, and 9.

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Gypsy
I thought this was a good one. All my friends who I've shown it to (except for one) haven't been able to solve it.

Many years from now, two men are sitting in the county park. The
following is part of their discussion:

Man 1: Yes, I'm married and have three fine sons.
Man 2: That's wonderful! How old are they?
MaN 1: Well, The product of their ages is equal to 36.
Man 2: Hmm. That doesn't tell me enough. Give me another clue.
Man 1: Ok, the sum of their ages is the number on that building
across the street.
Man 2: A ha! I've almost got the answer, but I still need another
clue.
Man 1: Very well. The oldest one has red hair.
Man 2: I've got it!
 
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in white:
2 2 9 [/color]
 
a nice puzzle

answer in white font 2, 2, and 9 ... there are 6 possible age combinations, only two give ambiguous sums of ages (2, 2, and 9; 1, 6, and 6) and it must be one of these or else the house number would have solved the problem in step 2; only 2, 2, and 9 gives a single 'oldest' child as the alternative (1, 6 and 6 has tiwn boys as oldest
 
Or simply 2, 3 and 6 years old.
 
I make it 9, 2 and 2
 
At first, that last clue seemed like something of a red hairing. :redface:
 
echoSwe said:
Or simply 2, 3 and 6 years old.
What other integer triplet with product 36 has sum 11 and also has the two largest members being equal to each other ?

Ha ha...a red hairing, indeed ! :smile:
 

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