Divisibility problem

The number $23$ is prime and divides every $23$ rd number. In all, there are $\lfloor \frac{2000}{23} \rfloor =86$ numbers from $1$ to $2000$ that are divisible by $23$ .

Among those $86$ numbers, three of them, namely $23^2, 2 \cdot 23^2, 3 \cdot 23^2$ are divisible by $23^2$ .

Hence $23^{89} | 2000!$ and $x=\boxed{89}$