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 age of the twins is 9 years while the age of the other is 7 years.
Edited on September 5, 2023, 9:33 pm
Puzzle Solution: Method II
