Let P stand for the product of the first 5 positive integer. What is the greatest possible value of m if `P/(10^m)` is an integer?

Options

A. 1
B. 2
C. 3
D. 5

Answer

To solve the problem, we need to find the greatest possible value of \( m \) such that \( \frac{P}{10^m} \) is an integer, where \( P \) is the product of the first 5 positive integers.

### Step-by-Step Solution:

1. **Calculate \( P \)**:
\[ P = 1 \times 2 \times 3 \times 4 \times 5 = 120 \]

2. **Express \( 10^m \)**:
\[ 10^m = 2^m \times 5^m \]
This means that for \( \frac{P}{10^m} \) to be an integer, \( P \) must have at least \( m \) factors of 2 and \( m \) factors of 5.

3. **Determine the prime factorization of \( P \)**:
\[ 120 = 2^3 \times 3^1 \times 5^1 \]
From this factorization, we see that \( P \) has 3 factors of 2 and 1 factor of 5.

4. **Find the limiting factor**:
- The number of factors of 2 in \( P \) is 3.
- The number of factors of 5 in \( P \) is 1.
- Since \( 10^m \) requires both \( 2^m \) and \( 5^m \), the limiting factor is the number of factors of 5.

5. **Determine the maximum value of \( m \)**:
- Since \( P \) has only 1 factor of 5, the maximum value of \( m \) can be at most 1.
- Therefore, \( m \) can be 0 or 1, but not more.

6. **Check if \( \frac{P}{10^m} \) is an integer for \( m = 1 \)**:
\[ \frac{120}{10^1} = \frac{120}{10} = 12 \]
This is an integer.

7. **Check if \( \frac{P}{10^m} \) is an integer for \( m = 2 \)**:
\[ \frac{120}{10^2} = \frac{120}{100} = 1.2 \]
This is not an integer.

8. **Conclusion**:
The greatest possible value of \( m \) such that \( \frac{P}{10^m} \) is an integer is:
\[ \boxed{1} \]