`((2007)!)/((2007)n)` is an integer & `n in N`, then find maximum value of n.

Answer

To solve the problem of finding the maximum value of \( n \) such that \(\frac{(2007)!}{(2000)^n}\) is an integer, we need to analyze the prime factorization of \( 2000 \) and how many times these prime factors appear in \( 2007! \).

### Step-by-Step Solution:

1. **Factorization of 2000**:
\[ 2000 = 2^4 \times 5^3 \]
This means that \( 2000^n = (2^4 \times 5^3)^n = 2^{4n} \times 5^{3n} \).

2. **Finding the number of factors of 2 in \( 2007! \)**:
We use the formula to find the number of times a prime \( p \) divides \( n! \):
\[ \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \]
For \( p = 2 \):
\[ \left\lfloor \frac{2007}{2} \right\rfloor + \left\lfloor \frac{2007}{4} \right\rfloor + \left\lfloor \frac{2007}{8} \right\rfloor + \left\lfloor \frac{2007}{16} \right\rfloor + \left\lfloor \frac{2007}{32} \right\rfloor + \left\lfloor \frac{2007}{64} \right\rfloor + \left\lfloor \frac{2007}{128} \right\rfloor + \left\lfloor \frac{2007}{256} \right\rfloor + \left\lfloor \frac{2007}{512} \right\rfloor + \left\lfloor \frac{2007}{1024} \right\rfloor \]
Calculating each term:
\[ \left\lfloor \frac{2007}{2} \right\rfloor = 1003, \quad \left\lfloor \frac{2007}{4} \right\rfloor = 501, \quad \left\lfloor \frac{2007}{8} \right\rfloor = 250 \]
\[ \left\lfloor \frac{2007}{16} \right\rfloor = 125, \quad \left\lfloor \frac{2007}{32} \right\rfloor = 62, \quad \left\lfloor \frac{2007}{64} \right\rfloor = 31 \]
\[ \left\lfloor \frac{2007}{128} \right\rfloor = 15, \quad \left\lfloor \frac{2007}{256} \right\rfloor = 7, \quad \left\lfloor \frac{2007}{512} \right\rfloor = 3, \quad \left\lfloor \frac{2007}{1024} \right\rfloor = 1 \]
Adding these values:
\[ 1003 + 501 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 1998 \]
Thus, the total number of factors of 2 in \( 2007! \) is \( 1998 \).

3. **Finding the number of factors of 5 in \( 2007! \)**:
For \( p = 5 \):
\[ \left\lfloor \frac{2007}{5} \right\rfloor + \left\lfloor \frac{2007}{25} \right\rfloor + \left\lfloor \frac{2007}{125} \right\rfloor + \left\lfloor \frac{2007}{625} \right\rfloor \]
Calculating each term:
\[ \left\lfloor \frac{2007}{5} \right\rfloor = 401, \quad \left\lfloor \frac{2007}{25} \right\rfloor = 80, \quad \left\lfloor \frac{2007}{125} \right\rfloor = 16, \quad \left\lfloor \frac{2007}{625} \right\rfloor = 3 \]
Adding these values:
\[ 401 + 80 + 16 + 3 = 500 \]
Thus, the total number of factors of 5 in \( 2007! \) is \( 500 \).

4. **Setting up the inequalities**:
We need:
\[ 4n \leq 1998 \quad \text{and} \quad 3n \leq 500 \]
From \( 4n \leq 1998 \):
\[ n \leq \frac{1998}{4} = 499.5 \quad \Rightarrow \quad n \leq 499 \]
From \( 3n \leq 500 \):
\[ n \leq \frac{500}{3} \approx 166.67 \quad \Rightarrow \quad n \leq 166 \]

5. **Conclusion**:
The maximum value of \( n \) that satisfies both inequalities is:
\[ n = 166 \]

### Final Answer:
The maximum value of \( n \) such that \(\frac{(2007)!}{(2000)^n}\) is an integer is \( \boxed{166} \).