All possible values of $x$

The highest power of a prime $p$ in the prime factorization of $n!$ is given by

$k = \sum_{i=1}^{\left\lfloor \log_{p} n \right\rfloor} \left\lfloor \dfrac{n}{p^i} \right\rfloor$

Since the number we want should be an integer, therefore we can take all values of $x$ in the above problem ranging from $0$ to $k$ , i.e., $0 \le x \le k$ .

Evaluating the value of $k$ , we get

$k = \sum_{i=1}^{\left\lfloor \log_{2} 13 \right\rfloor} \left\lfloor \dfrac{13}{2^i} \right\rfloor = \sum_{i=1}^{3} \left\lfloor \dfrac{13}{2^i} \right\rfloor = \left\lfloor \dfrac{13}{2} \right\rfloor + \left\lfloor \dfrac{13}{2^2} \right\rfloor + \left\lfloor \dfrac{13}{2^3} \right\rfloor = 6 + 3 + 1 = 10$

Thus, $\boxed{0 \le x \le 10}$ .