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Here is the answer from the website that I got the riddle from:
The most important feature to recognize in solving this puzzle is that Postman A needed to know that Postman B did not have twins in order to solve the puzzle. That means that there must be two possible solutions that satisfied the "age in years" and "number of windows" conditions of the problem. He needed to rule out twins to solve the problem. To summarize, there must have been two potential solutions with 3 son's ages having:
(i) equal products (the age condition)
(ii) equal sums (the windows condition)
(iii) one involving twins and one not.
If you really wanted to be fussy, since a house cannot have a fraction of a window, it also means that the ages of the sons must all be in integers (whole numbers) and not involve fractional ages.
The one solution to the problem that satisfies all of these conditions is (1, 5, 8) which has the same sum (14) and the same product (40) as another possible solution involving twins (2, 2, 10). There is just no other solution that works involving a postman's age less than 90 years old.
The possible alternative solution that you propose (2, 4, 6) does not work since the product (48) is different than 40 and the sum (12) is different than 14. In order for (2, 4, 6) to work, you would have to find another possible solution involving twins that sum to 12 and have 48 as a product. Two possibilities are:
(i) 2-2-12 (no; same product but different sum)
(ii) 3-3-6 (no; same sum but different product)
Cyberdog was right
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I am enough of an artist to draw freely upon my imagination. Imagination is more important than knowledge. Knowledge is limited. Imagination encircles the world.
Albert Einstein
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