Question

Number of prime factors in
$\frac{6^{12}\times (25{)}^{28}\times (15{)}^{16}}{(14{)}^{12}\times (21{)}^{11}}$

Answer 1

$=\frac{(2\times 3{)}^{12}\times (7\times 5{)}^{28}\times (3\times 5{)}^{16}}{(2\times 7{)}^{12}\times (3\times 7{)}^{11}}=\frac{2^{12}\times 3^{12}\times 7^{28}\times 5^{28}\times 3^{16}\times 5^{16}}{2^{12}\times 7^{12}\times 3^{11}\times 7^{11}}=2^{12-12}\times 3^{12+16-11}\times 7^{28-12-11}\times 5^{28+16}=2^0\times 3^{17}\times 7^5\times 5^{44}=3^{17}\times 7^5\times 5^{44}$
3 prime factors $\to 3,7\&5$