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

Let $d(n)$ denote the number of divisors of $n$ including $1$ and itself. Then,$d(225)$,$d(1125)$,and $d(640)$ are

Let $f(x) = 2x^n + \lambda$,where $\lambda \in R$ and $n \in N$. Given $f(4) = 133$ and $f(5) = 255$,find the sum of all the positive integer divisors of $(f(3) - f(2))$.

The largest $n \in \mathbb{N}$ such that $3^n$ divides $50!$ 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:

What is the highest power of $3$ in $100!$?