If 133! is divisible by ` 7^(n)` then find the maximum value of n.

Options

A. 21
B. 22
C. 23
D. 24

Answer

To find the maximum value of \( n \) such that \( 133! \) is divisible by \( 7^n \), we can use the formula for finding the highest power of a prime \( p \) that divides \( n! \):

\[ n_p(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \]

In our case, \( n = 133 \) and \( p = 7 \).

### Step-by-step Solution:

1. **Calculate \( \left\lfloor \frac{133}{7} \right\rfloor \)**:
\[ \frac{133}{7} = 19 \]
So, \( \left\lfloor \frac{133}{7} \right\rfloor = 19 \).

2. **Calculate \( \left\lfloor \frac{133}{7^2} \right\rfloor \)**:
\[ 7^2 = 49 \quad \text{and} \quad \frac{133}{49} \approx 2.714 \]
So, \( \left\lfloor \frac{133}{49} \right\rfloor = 2 \).

3. **Calculate \( \left\lfloor \frac{133}{7^3} \right\rfloor \)**:
\[ 7^3 = 343 \quad \text{and} \quad \frac{133}{343} < 1 \]
So, \( \left\lfloor \frac{133}{343} \right\rfloor = 0 \).

4. **Sum the results**:
Now, we add all the values we calculated:
\[ n = 19 + 2 + 0 = 21 \]

Thus, the maximum value of \( n \) such that \( 133! \) is divisible by \( 7^n \) is \( 21 \).

### Final Answer:
\[ n = 21 \]