Question

Total number of even divisors of ‘1323000’ which are divisible by 105 is $2^k$ – 10, then k is ___

Answer 1

$1323000=2^3{.3}^3{.5}^3{.7}^2$
For even divisor and divisible by 105, 2, 3, 5, 7 must occur at least one time.
$\therefore$ Total number of divisors are $3\times 3\times 3\times 2=54$
$Hence,2^k-10=54\Rightarrow k=6$