CAT 2015 Quantitative Ability

$\mathit{P}={15}^{100}\left(1\times 2\times 3\times ....\times 100\right)$=${15}^{100}\times 100!$
Highest power of 2 in P = 97 (2 will be deciding factor for number of zeroes because number of fives will be greater than number of zeroes in this number)
$\mathit{Q}={25}^{20+50}\left(1\times 2\times 3\mathit{x}..\times 50\right)=5^{2000}\times 50!$
Highest power of 2 in Q = 47
So Highest power of 2 in
$\frac{{\mathit{P}}^2}{\mathit{Q}}\times {10}^{1767}=2\times 97+1767-47=1914$
Hence, number of zeroes = 1914.