Find the number of zeros at the end of `1000!.`

Answer

To find the number of zeros at the end of \(1000!\), we need to determine how many times \(10\) is a factor in \(1000!\). Since \(10\) is the product of \(2\) and \(5\), and there are generally more factors of \(2\) than \(5\) in factorials, we only need to count the number of times \(5\) is a factor in \(1000!\).

### Step-by-Step Solution:

1. **Count the multiples of \(5\)**:
- To find how many multiples of \(5\) are there in \(1000\), we perform the division:
\[ \left\lfloor \frac{1000}{5} \right\rfloor = 200 \]
This means there are \(200\) numbers that contribute at least one factor of \(5\).

2. **Count the multiples of \(25\)**:
- Next, we count how many multiples of \(25\) are there in \(1000\):
\[ \left\lfloor \frac{1000}{25} \right\rfloor = 40 \]
These numbers contribute an additional factor of \(5\) each.

3. **Count the multiples of \(125\)**:
- Now, we count the multiples of \(125\):
\[ \left\lfloor \frac{1000}{125} \right\rfloor = 8 \]
Each of these contributes yet another factor of \(5\).

4. **Count the multiples of \(625\)**:
- Finally, we count the multiples of \(625\):
\[ \left\lfloor \frac{1000}{625} \right\rfloor = 1 \]
This contributes one more factor of \(5\).

5. **Sum all the factors of \(5\)**:
- Now, we add all these contributions together to find the total number of factors of \(5\) in \(1000!\):
\[ 200 + 40 + 8 + 1 = 249 \]

Thus, the number of zeros at the end of \(1000!\) is **249**.