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 The Aging Brothers (Posted on 2003-08-14)
The product of 3 brothers' ages is 567. Two are twins.

How old is the other one?

 See The Solution Submitted by Jayaram S Rating: 3.0000 (9 votes)

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 Puzzle Solution: Method II | Comment 17 of 18 |
(In reply to Puzzle Solution by K Sengupta)

Let the age of the twins be x years while the age of the other
sibling is y years

Case I: x>=y

x^2*y = 567(given)
-> x^3 >=  567
-> x>= 9

No square of x with x>=9, apart from x= 9, divides 567.

So, x=9, so that: y = 567/81 = 7

Case II: x < y

x^2*y = 567(given)
-> x^3 <  567
-> x <=8

The only possible cases are: x = 1, 3

If x = 1 -> y = 567/1 = 567, which is much greater than the age of a mere boy.

if x =3 -> y = 567/9 = 83, which is much greater than the age of a mere boy.

Consequently, comparing the two cases it follows that the other one is 7 years old.

Edited on December 13, 2008, 12:32 pm
 Posted by K Sengupta on 2008-12-13 12:24:00

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