Find the number of Zeros at the end of 378!

Options

A. 93
B. 90
C. 75
D. 81

Answer

To find the number of zeros at the end of 378!, we need to determine how many times 10 is a factor in the factorial. Since 10 is made up of the factors 2 and 5, and there are generally more factors of 2 than factors of 5 in factorials, we focus on counting the number of times 5 is a factor in 378!.

### Step-by-Step Solution:

1. **Identify the formula**: The number of trailing zeros in a factorial can be found using the formula:
\[ \text{Number of trailing zeros} = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{5^2} \right\rfloor + \left\lfloor \frac{n}{5^3} \right\rfloor + \ldots \]
where \( n \) is the number for which we are calculating the factorial.

2. **Calculate the first term**: For \( n = 378 \):
\[ \left\lfloor \frac{378}{5} \right\rfloor = \left\lfloor 75.6 \right\rfloor = 75 \]

3. **Calculate the second term**: Next, we calculate:
\[ \left\lfloor \frac{378}{25} \right\rfloor = \left\lfloor 15.12 \right\rfloor = 15 \]

4. **Calculate the third term**: Now, we calculate:
\[ \left\lfloor \frac{378}{125} \right\rfloor = \left\lfloor 3.024 \right\rfloor = 3 \]

5. **Calculate the fourth term**: Finally, we check for the next power of 5:
\[ \left\lfloor \frac{378}{625} \right\rfloor = \left\lfloor 0.6048 \right\rfloor = 0 \]

6. **Sum all the terms**: Now, we add all the values we calculated:
\[ 75 + 15 + 3 + 0 = 93 \]

Thus, the number of trailing zeros in 378! is **93**.