How many natural-number factors does N have if N=2^4*3^8*5^2*7^2?
2^4*3^8*5^2*7^2==2 , 2 , 2 , 2 , 3 , 3 , 3 , 3 , 3 , 3 , 3 , 3 , 5 , 5 , 7 , 7>> Total Prime Factors = 16
[4+1] x [8+1] x [2+1] x [2+1]==405 divisors.