Question

What is the total number of factors of the number $N=4^{11}\times {14}^5\times {11}^2$ ?

Answer 1

$N=4^{11}\times {14}^5\times {11}^2=(2^2{)}^{11}\times (2\times 7{)}^5\times {11}^2=2^{22}\times 2^5\times 7^5\times {11}^2=2^{27}\times 7^5\times {11}^2$
$\therefore$ Total number of prime factors of
$N=(27+1)\times (5+1)\times (2+1)=28\times 6\times 3=504$