Numbers 2, 4, 6, 8, 10, ...., 196, 198, 200 are multiplied together. The number of zeros at the end of the product on the right will be equal to -

Options

A. 21
B. 22
C. 24
D. 25

Answer

To find the number of zeros at the end of the product of the numbers 2, 4, 6, ..., 200, we need to determine how many times the factors 2 and 5 appear in the product, as each pair of these factors contributes one trailing zero.

### Step-by-Step Solution:

1. **Identify the Sequence**:
The numbers we are multiplying are all even numbers from 2 to 200. This can be expressed as:
\[ 2, 4, 6, 8, \ldots, 200 \]
This is an arithmetic sequence where the first term \(a = 2\) and the last term \(l = 200\) with a common difference \(d = 2\).

2. **Count the Total Numbers**:
To find the number of terms in this sequence, we can use the formula for the \(n\)-th term of an arithmetic sequence:
\[ n = \frac{l - a}{d} + 1 = \frac{200 - 2}{2} + 1 = 100 \]
So, there are 100 terms.

3. **Count the Factors of 2**:
Since all numbers are even, we can factor out 2 from each term:
\[ 2 \times (1, 2, 3, \ldots, 100) \]
This means we have at least 100 factors of 2. Now, we need to count additional factors of 2 from the multiples of 4, 8, 16, etc.

- Count multiples of \(2\): \(100\)
- Count multiples of \(4\): \(\left\lfloor \frac{100}{2} \right\rfloor = 50\)
- Count multiples of \(8\): \(\left\lfloor \frac{100}{4} \right\rfloor = 25\)
- Count multiples of \(16\): \(\left\lfloor \frac{100}{8} \right\rfloor = 12\)
- Count multiples of \(32\): \(\left\lfloor \frac{100}{16} \right\rfloor = 6\)
- Count multiples of \(64\): \(\left\lfloor \frac{100}{32} \right\rfloor = 3\)
- Count multiples of \(128\): \(\left\lfloor \frac{100}{64} \right\rfloor = 1\)

Adding these up gives:
\[ 100 + 50 + 25 + 12 + 6 + 3 + 1 = 197 \]
So, there are 197 factors of 2.

4. **Count the Factors of 5**:
Now, we need to count the factors of 5 in the product:
- Count multiples of \(5\): \(\left\lfloor \frac{100}{5} \right\rfloor = 20\)
- Count multiples of \(25\): \(\left\lfloor \frac{100}{25} \right\rfloor = 4\)

Adding these gives:
\[ 20 + 4 = 24 \]
So, there are 24 factors of 5.

5. **Determine the Number of Trailing Zeros**:
The number of trailing zeros in the product is determined by the minimum of the counts of factors of 2 and 5:
\[ \text{Trailing Zeros} = \min(197, 24) = 24 \]

### Final Answer:
The number of zeros at the end of the product is **24**.