The total number of $3$-digit numbers whose greatest common divisor (g.c.d.) with $36$ is $2$ is:

The largest non-negative integer $k$ such that $24^k$ divides $13!$ is

Consider the following statements:
$I$: The number of non-trivial even divisors of the number $N = 2^{\alpha_1} 3^{\alpha_2} 4^{\alpha_3} 5^{\alpha_4} 6^{\alpha_5}$ is $(\alpha_1+2\alpha_3+\alpha_5)(\alpha_2+\alpha_5+1)(\alpha_4+1)-1$.
$II$: The number of non-trivial odd divisors of the number $N = 2^{\alpha_1} 3^{\alpha_2} 4^{\alpha_3} 5^{\alpha_4} 6^{\alpha_5}$ is $\alpha_2+\alpha_4+\alpha_5+\alpha_2\alpha_4+\alpha_4\alpha_5$. Then:

The number $111...1$ ($91$ times) is

The number of zeros at the end of $100!$ is

If $p$ is a prime number,then $n^p - n$ is divisible by $p$ when $n$ is a