Given two numbers $n$ and $k$ find the greatest power of $k$ that divides $n!$

Writing the factorial expression explicitly

We can see that every $k$-th member of the factorial is divisible by $k$; therefore, one answer is $\lfloor \frac{n}{k}\rfloor$. However, we can also see that every $k^2$-th term is also divisible by $k$ two times, and it gives one more term to the answer; that is, $\lfloor \frac{n}{k^2}\rfloor$, which means that every $k^i$-th term adds one factor to the answer; thus, the answer is

The sum is actually finite, and the maximum value of $i$ can be found using logarithms. Let $k^i>n$. Applying logarithms, we have $i\cdot log(k)>log(n)$, which is equal to $i>\frac{log(n)}{log(k)}$, which is the same as $i>lo{g}_{k}n$.

The sum discovered by Adrien-Marie Legendre is called Legendre’s Formula . Let $d_a(b)$ be the number of times $a$ divides $b$

/**
 * Computes the maximum power of `k` that is a divisor of `n!`
 *
 * @param {int} n
 * @param {int} k
 * @return {int}
 */
int max_power_in_factorial(int n, int k) {
  int ans = 0;
  while (n) {
    n /= k;
    ans += n;
  }
  return ans;
}