Question

How many three digit numbers with distinct digits can be formed such that product of the digits is the cube of a positive integer?
a)21 b)24 c)36 d)30

Answer 1

Let one digit is fixed to be 1. Then product of two distinct digits must be a perfect cube i.e.
(2,4) such that 1*2*4 = 8 = 23 total 3! = 6 combinations
(3,9) such that 1*3*9 = 27 = 33 total 3! = 6 combinations
One digit is fixed to be 2.
(4,8) such that 2*4*8 = 64 = 43 total 3! = 6 combinations
One digit is fixed to be 3.
(8,9) such that 3*8*9 = 216 = 63 total 3! = 6 combinations
One digit is fixed to be 4
(6,9) such that 4*6*9 = 216 = 63 total 3! = 6 combinations

Answer 2

thanks @zpupster

Answer 3

@zpupster I got you except 23 total, 33 total, 43 total.... what do they mean, please explain me.
btw , your explanation is perfect