The number 1, 2, 3, 4, ...., 1000 are multiplied together. The number of zeros at the end (on the right) of the product must be:

Options

A. 30
B. 200
C. 211
D. 249

Answer

To find the number of zeros at the end of the product of the numbers from 1 to 1000 (which is \(1000!\)), we need to determine how many times 10 is a factor in this product. Since \(10 = 2 \times 5\), we need to find the pairs of factors of 2 and 5 in \(1000!\).

However, in any factorial, the number of factors of 2 will always be greater than the number of factors of 5. Therefore, the number of zeros at the end of \(1000!\) will be determined by the number of times 5 is a factor in the numbers from 1 to 1000.

### Step-by-step Solution:

1. **Identify the Formula**: The number of times a prime \(p\) is a factor in \(n!\) can be calculated using the formula:
\[ \text{Number of } p \text{ in } n! = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots \]
until \(p^k > n\).

2. **Set \(n = 1000\) and \(p = 5\)**: We will calculate the number of times 5 is a factor in \(1000!\).

3. **Calculate Each Term**:
- First term:
\[ \left\lfloor \frac{1000}{5} \right\rfloor = \left\lfloor 200 \right\rfloor = 200 \]
- Second term:
\[ \left\lfloor \frac{1000}{25} \right\rfloor = \left\lfloor 40 \right\rfloor = 40 \]
- Third term:
\[ \left\lfloor \frac{1000}{125} \right\rfloor = \left\lfloor 8 \right\rfloor = 8 \]
- Fourth term:
\[ \left\lfloor \frac{1000}{625} \right\rfloor = \left\lfloor 1.6 \right\rfloor = 1 \]
- Fifth term:
\[ \left\lfloor \frac{1000}{3125} \right\rfloor = \left\lfloor 0.32 \right\rfloor = 0 \]

4. **Sum the Results**:
Now, we add all the results together:
\[ 200 + 40 + 8 + 1 + 0 = 249 \]

5. **Conclusion**:
Therefore, the number of zeros at the end of the product of the numbers from 1 to 1000 is **249**.