Question

Number of prime factors in
$\frac{6^{12}\times (35{)}^{28}\times (15{)}^{16}}{(14{)}^{12}\times (21{)}^{11}}$

Answer 1

$\frac{6^{12}\times (35{)}^{28}\times (15{)}^{16}}{(14{)}^{12}\times (21{)}^{11}}$
$=\frac{(3{)}^{12}\times (2{)}^{12}\times \left(7\right){\times }^{28}\times (5{)}^{28}\times (5{)}^{16}\times (3{)}^{16}}{(7{)}^{12}\times (2{)}^{12}\times (3{)}^{11}\times (7{)}^{11}}$
$=(3{)}^{17}\times (5{)}^{44}\times (7{)}^5$
$\therefore$ Required number of prime factors
$=17+44+5=66$