What is the smallest of the eight natural numbers with different divisors?
The number of divisors is equal to the product of prime factor times plus 1
8= (1+1)(1+1)(1+1) = (1+1)(3+1) = (7+1)
So the form of n is p * Q * r, or p * q ^ 3, or P ^ 7
The minimum numbers corresponding to the above forms are:
2*3*5 = 30
2^3 * 3 = 24
2^7 = 128
So the smallest number is 24