1. **Problem statement:** Find the greatest integer $q$ such that $4^q$ divides the product $p = 1 \times 2 \times 3 \times \cdots \times 1000$, which is $1000!$. 2. **Understanding the problem:** Since $4 = 2^2$, finding the highest power of $4$ dividing $1000!$ is equivalent to finding the highest power of $2^{2q}$ dividing $1000!$. Thus, we need to find the exponent of 2 in the prime factorization of $1000!$ and then divide it by 2. 3. **Formula for the exponent of a prime $p$ in $n!$:** $$ \text{exponent of } p \text{ in } n! = \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor $$ 4. **Calculate the exponent of 2 in $1000!$:** $$ \left\lfloor \frac{1000}{2} \right\rfloor + \left\lfloor \frac{1000}{2^2} \right\rfloor + \left\lfloor \frac{1000}{2^3} \right\rfloor + \left\lfloor \frac{1000}{2^4} \right\rfloor + \left\lfloor \frac{1000}{2^5} \right\rfloor + \left\lfloor \frac{1000}{2^6} \right\rfloor + \left\lfloor \frac{1000}{2^7} \right\rfloor + \left\lfloor \frac{1000}{2^8} \right\rfloor + \left\lfloor \frac{1000}{2^9} \right\rfloor + \left\lfloor \frac{1000}{2^{10}} \right\rfloor $$ Calculate each term: - $\left\lfloor \frac{1000}{2} \right\rfloor = 500$ - $\left\lfloor \frac{1000}{4} \right\rfloor = 250$ - $\left\lfloor \frac{1000}{8} \right\rfloor = 125$ - $\left\lfloor \frac{1000}{16} \right\rfloor = 62$ - $\left\lfloor \frac{1000}{32} \right\rfloor = 31$ - $\left\lfloor \frac{1000}{64} \right\rfloor = 15$ - $\left\lfloor \frac{1000}{128} \right\rfloor = 7$ - $\left\lfloor \frac{1000}{256} \right\rfloor = 3$ - $\left\lfloor \frac{1000}{512} \right\rfloor = 1$ - $\left\lfloor \frac{1000}{1024} \right\rfloor = 0$ 5. **Sum these values:** $$ 500 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 + 0 = 994 $$ 6. **Find $q$ such that $4^q = 2^{2q}$ divides $1000!$:** Since the exponent of 2 in $1000!$ is 994, the greatest $q$ satisfies: $$ 2q \leq 994 \implies q = \left\lfloor \frac{994}{2} \right\rfloor = 497 $$ **Final answer:** $\boxed{497}$ (Option D)