The digit at Hundred's place value of 17! is

Options

A. 1
B. 0
C. 2
D. 3

Answer

To find the digit at the hundred's place value of \( 17! \), we first need to understand how many trailing zeros are in \( 17! \). The number of trailing zeros in a factorial can be determined by counting the number of times 5 is a factor in the numbers from 1 to 17. This is because there are usually more factors of 2 than factors of 5, and each pair of 2 and 5 contributes to a trailing zero.

### Step-by-Step Solution:

1. **Calculate the number of trailing zeros in \( 17! \)**:
- The formula to find the number of trailing zeros in \( n! \) is:
\[ \text{Number of trailing zeros} = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \ldots \]
- For \( n = 17 \):
- Calculate \( \left\lfloor \frac{17}{5} \right\rfloor = 3 \)
- Calculate \( \left\lfloor \frac{17}{25} \right\rfloor = 0 \) (since 25 > 17)
- Therefore, the total number of trailing zeros in \( 17! \) is:
\[ 3 + 0 = 3 \]

2. **Interpret the result**:
- The presence of 3 trailing zeros means that \( 17! \) ends with three zeros. This means that the last three digits of \( 17! \) are 000.

3. **Determine the digit at the hundred's place**:
- Since \( 17! \) ends in three zeros, the digit at the hundred's place (which is the third digit from the right) is 0.

### Final Answer:
The digit at the hundred's place value of \( 17! \) is **0**.