If No . 1 to 50 are multiplied together . Find the max power of 7.

Answer

To find the maximum power of 7 in the product of numbers from 1 to 50 (which is represented as 50!), we can use the formula for finding the highest power of a prime \( p \) that divides \( n! \):

\[ \text{Maximum power of } p \text{ in } n! = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots \]

Where \( \left\lfloor x \right\rfloor \) denotes the greatest integer less than or equal to \( x \).

### Step-by-Step Solution:

1. **Identify \( n \) and \( p \)**:
- Here, \( n = 50 \) and \( p = 7 \).

2. **Calculate \( \left\lfloor \frac{50}{7} \right\rfloor \)**:
- \( \frac{50}{7} \approx 7.14 \)
- Thus, \( \left\lfloor \frac{50}{7} \right\rfloor = 7 \).

3. **Calculate \( \left\lfloor \frac{50}{7^2} \right\rfloor \)**:
- \( 7^2 = 49 \)
- \( \frac{50}{49} \approx 1.02 \)
- Thus, \( \left\lfloor \frac{50}{49} \right\rfloor = 1 \).

4. **Calculate \( \left\lfloor \frac{50}{7^3} \right\rfloor \)**:
- \( 7^3 = 343 \)
- \( \frac{50}{343} \approx 0.15 \)
- Thus, \( \left\lfloor \frac{50}{343} \right\rfloor = 0 \).

5. **Sum the results**:
- Now, we add the results from the above calculations:
\[ 7 + 1 + 0 = 8 \]

### Conclusion:
The maximum power of 7 that divides \( 50! \) is **8**.