The product of 3 brothers' ages is 567. Two are twins.

How old is the other one?

(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**