Smallest n

$n! = \displaystyle \prod_{k=1}^m p_k^{q_k}$ , the $n$ factorial is the product of powers $q_k$ of all primes $p_k$ smaller of equal to $n$ . The power of the $k$ th prime is given by $\displaystyle q_k = \sum_{j=1}^\infty \left \lfloor \frac n{p_k^j}\right \rfloor$ , where $\lfloor \cdot \rfloor$ denotes the floor function . Denote $q_k = q(p_k)$ . Then we have:

$\begin{aligned} q(2) & = \sum_{j=1}^\infty \left \lfloor \frac {50}{2^j} \right \rfloor \\ & = \left \lfloor \frac {50}2 \right \rfloor + \left \lfloor \frac {50}4 \right \rfloor + \left \lfloor \frac {50}8 \right \rfloor + \left \lfloor \frac {50}{16} \right \rfloor + \left \lfloor \frac {50}{32} \right \rfloor \\ & = 25 + 12 + 6 + 3 + 1 = 47 \end{aligned}$

$\begin{aligned} q(3) & = \sum_{j=1}^\infty \left \lfloor \frac {50}{3^j} \right \rfloor \\ & = \left \lfloor \frac {50}3 \right \rfloor + \left \lfloor \frac {50}9 \right \rfloor + \left \lfloor \frac {50}{27} \right \rfloor \\ & = 16 + 5 + 1 = 22 \end{aligned}$

Therefore, $\dfrac {n!}{24^n} = \dfrac {2^{47}\cdot 3^{22}\cdot 5^{12}\cdots 43\cdot 47}{2^{3n}3^n}$ is not an integer, when $24^n \not \mid 50!$ or $3n > 47$ or $n > 22$ . $\implies n \ge \boxed{16}$ .

Mathematica

Min@Select[Range@100,!IntegerQ[50!/24^#]&]

returns *16 *