Can't think of a good name right now

Note that 24 = 2^3×3 So we need to find the powers of 2 and 3 in 50! so that we can find the least power of 24 which exceeds that value. The power of 2 in 50! can be calculated as such: floor(50/2) = 25, floor(25/2) = 12, floor(12/2) = 6, floor(6/2) = 3, floor(3/2) = 1. After adding them up, we get 2^47, whose power is just one short of 48. For power of 3, we can use the same method to get 3^22. Notice that the ratio of 2 to 3 in 24 is obviously larger than that in 50!. Therefore, we only need to find the largest power of 2 exceeding 47 and divisible by 3 (so that n is an integer). In this case, it is 48. Therefore, we need 16 24s in order to exceed the power of 2 in 50!. Thus n = 16

Let $A = \frac{50!}{24^n}$

$\Rightarrow A = \frac{50!}{2^{3n} \cdot 3^n }$

We need to find $a,b$ such that $50! = 2^{a} \cdot 3^{b} \cdot (...)$ ( the $(...)$ signifies the prime factors that we don't need to care about )

We utilise Legendre's formula for the $2,3$ -adic valuation:

$a = \nu_{2}(50!) = \sum _{ i=1 }^{ \infty }{ \left\lfloor \frac { 50 }{ 2^{ i } } \right\rfloor } = 47$

$b = \nu_{3}(50!) = \sum _{ i=1 }^{ \infty }{ \left\lfloor \frac { 50 }{ 3^{ i } } \right\rfloor } = 22$

$\Rightarrow A = \frac{50!}{2^{3n} \cdot 3^n } = \frac{2^{47}\cdot 3^{22} \cdot (...)}{2^{3n} \cdot 3^n}$

$\Leftrightarrow A = 2^{47-3n}\cdot 3^{22-n} \cdot (...)$

For $A \notin \mathbb{Z}$ , we have 2 cases: either the power of the prime factor $2$ or of the prime factor $3$ , is less than 0. The purpose is to attain those powers to negative value because the powers will make $A$ a fraction, therefore not an integer.

Case 1: $47-3n \lt 0$

$\Leftrightarrow n \gt 15.\overline { 6 }$ , hence we can safely let $\boxed{n = 16}$

Case 2: $22-n \lt 0$

$\Leftrightarrow \boxed{n \gt 22}$

We can see that Case 1 is the minimal solution to $n$

Hence, the smallest positive integer that makes $A \notin \mathbb{Z}$ is $\boxed{n = 16}$