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Mathematician A asks Mathematician B to guess the age of his three sons. Mathematician A starts off by giving Mathematician B two clues. The two clues are:

  • The product of their ages is 72.
  • When you add the ages of the sons, it will be the same as the number of apples on his his apple tree.

After these two clues, Mathematician B claims that it is impossible to find the ages. And so, Mathematician A gives yet one more clue.

  • The eldest son loves to play basketball.

After this clue, Mathematician B claims that he has found the ages of the sons.

How old are Mathematician A's sons?

  • Find all possible age combinations, and for each find the sum. Staring at the numbers may produce an epiphany. – André Nicolas May 29 '15 at 23:38
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    what did you try? did you at least start off by decomposing $72$ as the product of three integers? Did you then consider the possible sums? Did you go through some of those possibilities? did you eliminate impossible ones. Did you look at what's left? Maybe then the final clue will make sense to you. – Ittay Weiss May 29 '15 at 23:39
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    This relies on the (rather ridiculous) assumption that if two people have the same age (as an integer), there wouldn't be an "eldest". – Robert Israel May 29 '15 at 23:39
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    Does mathematician $B$ know the number of apples in mathematician $A$'s apple tree? – Asinomás May 29 '15 at 23:40
  • Indeed. In language, even in the extreme case, you can have the "older twin" which may be older by only a few seconds or minutes. – JMoravitz May 29 '15 at 23:43
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    Yes, she knows, or at least can find out, the number of apples. If it could be done by counting apples it would not be "impossible." – André Nicolas May 29 '15 at 23:46
  • @RobertIsrael: I assumed that the ages were integers and it worked out pretty well. So I think OP wants us to assume that the ages are integers. –  May 29 '15 at 23:46
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    Also http://math.stackexchange.com/questions/581225/what-are-the-ages, and see similar http://math.stackexchange.com/questions/40158/logic-puzzle-of-the-age-of-three-sons – Chris Culter May 29 '15 at 23:53
  • @RobertIsrael For children of the same person the assumption isn't that ridiculous. –  May 30 '15 at 00:19
  • Even with twins, one is older (maybe only by a few minutes). – Robert Israel May 30 '15 at 02:44

1 Answers1

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Let us first take a look at the first clue: "The product of their ages is $72$." This means that there are only $12$ possible sets of ages: $$\begin{array}{|c|c|c|} \hline \text{Age of 1st Son} & \text{Age of 2nd Son} & \text{Age of 3rd Son}\\ \hline 1 & 1 & 72\\ 1 & 2 & 36\\ 1 & 3 & 24\\ 1 & 4 & 18\\ 1 & 6 & 12\\ 1 & 8 & 9\\ 2 & 2 & 18\\ 2 & 3 & 12\\ 2 & 4 & 9\\ 2 & 6 & 6\\ 3 & 3 & 8\\ 3 & 4 & 6\\ \hline \end{array}$$

Now let's look at the second clue: "When you add the ages of the sons, it will be the same as the number of apples on his his apple tree." Let us start by taking the sum of all the ages: $$\begin{array}{|c|c|c|c|} \hline \text{Age of 1st Son} & \text{Age of 2nd Son} & \text{Age of 3rd Son} & \text{Sum of 3 Sons}\\ \hline 1 & 1 & 72 & 74\\ 1 & 2 & 36 & 39\\ 1 & 3 & 24 & 28\\ 1 & 4 & 18 & 23\\ 1 & 6 & 12 & 19\\ 1 & 8 & 9 & 18\\ 2 & 2 & 18 & 22\\ 2 & 3 & 12 & 17\\ 2 & 4 & 9 & 15\\ \color{red}2 & \color{red}6 & \color{red}6 & \color{red}{14}\\ \color{red}{3} & \color{red}3 & \color{red}8 & \color{red}{14}\\ 3 & 4 & 6 & 13\\ \hline \end{array}$$

From this, notice that both $(2,6,6)$ and $(3,3,8)$ both add up to $14$. This is where the clue comes into play. We know that one of these two sets is what we are looking for, since Mathematician $B$ cannot find the ages from counting the apples on the apple tree.

Now that we are left with two possible sets, we look to our third clue. The third clue states: "The eldest son loves to play basketball." This means that there is only one elder son, and that there cannot be two identical large numbers in our set. Therefore we can eliminate the set $(2,6,6)$ and are left with $(3,3,8)$.

This means that the ages of the Mathematician $A$'s sons are $3$ years old, $3$ years old, and $8$ years old.