Prime Factor $^{x}$

Let $p$ be the largest prime factor of $\frac{1000!}{900!}$ that has an exponent larget than $1$ .

We need at least $2$ multiples of $p$ to be between $900$ and $1000$ (inclusive)

If $p>100$ , then there can be only 1 multiple of $p$ between $900$ and $1000$ . Example: if $p=101$ , then the only multiple of $p$ between $900$ and $1000$ is $909$ . The next multiple is $1010$ which is greater $1000$ .

The largest prime less than $100$ is $97$ . However, the only multiple of $97$ between $900$ and $1000$ is $970$ .

The next largest prime less than $100$ is $89$ . Again, there is only one multiple of $89$ between $900$ and $1000$ is $979$ .

The next largest prime less than $100$ is $83$ . This time, there are two multiples of $83$ between $900$ and $1000$ : $913$ and $996$ .

Thus the prime factorization of $\frac{1000!}{900!}$ would be $... \cdot 89^{2} \cdot ...$ with $89$ being the largest prime with a power of $2$

So the answer is $\boxed{83}$