How Did the Blind Mathematician Solve the King's Puzzle?

[AKG]

Once upon a time a king wanted to hire the best Mathematician in his kingdom to work in his palace. His servants brought to him the best two mathematicians, one of them was blind. The king told both mathematicians that he can't hire both so he will ask a question and whoever answers gets the job.

The king said: "I have three sons, whoever guesses their ages will be hired." The king told them that if they multiply the ages of his sons the result will be 36. Both mathematicians told the king that this information is not enough. The king then said: "The number of windows in the building across the street is equal to the sum of the ages of my sons." The first mathematician (who can see) counted the windows, and told the king that he still could not figure it out. The Blind mathematician (who could not count the windows) told the king that he does not have an answer. The king then said: "My oldest son has red hair." Right away the blind mathematician gave the correct answer and got hired.

The question is: How did the Blind guy know the answer and what is the answer?

(http://www.cs.uic.edu/~wsunna/thoughts/puzzle.html" )

 

[siddharth]

I think the answer is 9,2,2

My reasoning is as follows

Since the 3 ages are whole numbers the possible ages of the 3 sons are finite.
For example (6,2,3) (4,3,3) (6,6,1) (9,4,1) (18,2,1) (36,1,1) (12,3,1) (9,2,2)
Now since the sum of the three ages is given, one can eliminate some of the answers. The fact that, the mathematician who can see, counted the number of windows and still was not able to pick the right answer means that there are 2 sets which have the same sum. Namely (9,2,2) and (6,6,1).
Once the king told that his eldest son has red hair, the choice has to be (9,2,2) because there can be only 1 eldest son.

 

[honestrosewater]

Assuming their ages are whole numbers, knowing that there was an oldest son determined the answer, so for some unique number of windows w, there is a solution set to

abc = 36
a + b + c = w

that has more than one member, only one of which has a greatest element. It's w = 13: {(9, 2, 2), (6, 6, 1)}. Their ages are 9, 2, and 2. ??

 

[AKG]

You're both right.

 

[nate808]

the only problem with this is that logically there is an oldest son in both cases. For example, twins can both be born on the same day but one is always older. Did I miss something or am I just being too overanalytical?

 

[AKG]

Well, you're not being too overanalytical, that's an overstatement. But I can't decide whether you're being too analytical, or simply overanalytical. Also, it's possible that this king has two wives, and in the case where the sons' ages are 6, 6, and 1, both wives gave birth to the children that are now 6-years-old at the exact same time.

You were, of course, to assume that the ages were whole numbers, and in doing so, differences in age on the order of a few minutes would not count as making one actually older than the other.

 

[mathwonk]

well it could be that hair is not red on young children and only changes to red at a certain age.

 

[LarrrSDonald]

I've pondered the same "overanalytical" thing before when faced with similar questions. It certainly doesn't take much to realize where the problem means to aim with it, but it is a little dubious. Even with one wife, {6,6,1} could be a posibillity, in fact with the way my sisters and I were born (to one mother and one father) we were at one point {6,6,1}, if only for a little over a month. They were less then a year apart (me being five behind), making the younger turn six before the other turned 7 and both before I turned 2. The oldest was still considered the oldest, even during this "integer identical" period. It only falls apart because she doesn't have red hair, nor was she a son :-).

 

[Ubern0va]

Good one AKG. Fun to solve :)