Problem 1323

To find the greatest positive integer $x$ such that $23^{6+x}$ divides $2000!$, we need to determine how many times the prime factor 23 appears in the prime factorization of $2000!$.

The exponent of a prime $p$ in $n!$ is given by:
$$\sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor$$
In this case, $n = 2000$ and $p = 23$.

Let's calculate each term until we reach a power where the division results in a number less than 1:

1. $\left\lfloor \frac{2000}{23} \right\rfloor = \left\lfloor 86.95652 \right\rfloor = 86$

2. $\left\lfloor \frac{2000}{23^2} \right\rfloor = \left\lfloor \frac{2000}{529} \right\rfloor = \left\lfloor 3.78337 \right\rfloor = 3$

3. $\left\lfloor \frac{2000}{23^3} \right\rfloor = \left\lfloor \frac{2000}{12167} \right\rfloor = \left\lfloor 0.16438 \right\rfloor = 0$

For higher powers of 23, such as $23^4$, the floor function results in 0, since $23^4 = 279841$ is greater than 2000.

Therefore, the total number of times 23 appears as a factor in $2000!$ is:
$$86 + 3 + 0 = 89$$

We want $23^{6+x}$ to divide $2000!$, so we set:
$$6 + x \leq 89$$
Solving for $x$, we get:
$$x \leq 83$$

Therefore, the greatest positive integer $x$ is:
$$\boxed{83}$$