Find the number of naughts standing at the end of 125! .

Options

A. 29
B. 30
C. 31
D. 32

Answer

To find the number of trailing zeros in \( 125! \), we can use the formula that counts the number of times 5 is a factor in the numbers from 1 to 125. This is because the number of trailing zeros is determined by the pairs of factors of 2 and 5, and there are always more factors of 2 than factors of 5 in factorials.

### Step-by-Step Solution:

1. **Identify the formula**: The number of trailing zeros in \( n! \) can be calculated using the formula:
\[ Z(n) = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{5^2} \right\rfloor + \left\lfloor \frac{n}{5^3} \right\rfloor + \ldots \]
where \( Z(n) \) is the number of trailing zeros in \( n! \).

2. **Substitute \( n = 125 \)**:
\[ Z(125) = \left\lfloor \frac{125}{5} \right\rfloor + \left\lfloor \frac{125}{5^2} \right\rfloor + \left\lfloor \frac{125}{5^3} \right\rfloor \]

3. **Calculate each term**:
- First term:
\[ \left\lfloor \frac{125}{5} \right\rfloor = \left\lfloor 25 \right\rfloor = 25 \]
- Second term:
\[ \left\lfloor \frac{125}{25} \right\rfloor = \left\lfloor 5 \right\rfloor = 5 \]
- Third term:
\[ \left\lfloor \frac{125}{125} \right\rfloor = \left\lfloor 1 \right\rfloor = 1 \]
- Fourth term:
\[ \left\lfloor \frac{125}{625} \right\rfloor = \left\lfloor 0.2 \right\rfloor = 0 \]
(We stop here since further terms will also be 0.)

4. **Sum the results**:
\[ Z(125) = 25 + 5 + 1 = 31 \]

5. **Final answer**:
The number of trailing zeros in \( 125! \) is \( 31 \).