A conversation took place between two friends, a philosopher and a mathematician, who had not seen each or heard from each other in years. The mathematician, who had an exceedingly good memory, asked the philosopher how many children he had. The philosopher replied that he had three. The mathematician then asked how old the children were. His friend, who knew how much most mathematicians enjoy puzzles, said that he would give him a number of clues to the children's ages.
The philosopher's first clue: "The product of the children's ages is 36." The mathematician immediately replied that this was insufficient information.
The philosopher's second clue: "All of the children's ages are integers; none are fractional ages, such as 1 and ¼ years old." Still, the mathematician could not deduce the correct answer.
The philosopher's third clue: "The sum of the three children's ages is identical to the address of the house where we played chess together often, years ago." The mathematician still required more information (but assume the mathematician remembered the street number).
The philosopher then gave his fourth clue: "The oldest child looks like me." At this point, the mathematician was able to determine the ages of the three children. Here is your problem: What are the ages of the three children?
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The children's ages are 9,2,2.
ReplyDeleteSolution:
The first and second clues allow you to decompose 36 into its prime factors and begin recombining them into triples. This gives a list of them - can post later.
With the third clue, you can pick two of the groups - 2 2 9 and 1 6 6. This is because of all the multiples, only those two add up to the same number (13) - which is why the mathematician could not solve the problem.
Clue number 4 tells you that there is one oldest child which eliminates 1, 6, 6 leaving 2, 2, 9 as the solution.
:-)