The S={(1,3,5,7,9,…,99)(102,104,106,…200)} i.e., in the first part there are odd integers less than 100 and in the second part there are even integers greater than 100, but upto 200.
The highest power of 5 that can exactly divide the product is

Options

A. 25
B. 24
C. 30
D. 26

Answer

To find the highest power of 5 that can exactly divide the product of the set \( S = \{(1, 3, 5, 7, 9, \ldots, 99), (102, 104, 106, \ldots, 200)\} \), we will analyze both parts of the set separately.

### Step 1: Analyze the first part of the set
The first part consists of odd integers less than 100. These integers can be represented as:
\[ 1, 3, 5, 7, 9, \ldots, 99 \]

This is an arithmetic progression (AP) where:
- First term \( a = 1 \)
- Common difference \( d = 2 \)
- Last term \( l = 99 \)

To find the number of terms \( n \) in this AP, we use the formula:
\[ n = \frac{l - a}{d} + 1 \]
Substituting the values:
\[ n = \frac{99 - 1}{2} + 1 = \frac{98}{2} + 1 = 49 + 1 = 50 \]
So, there are 50 odd integers less than 100.

### Step 2: Count multiples of 5 in the first part
Next, we need to find how many of these integers are multiples of 5. The odd multiples of 5 less than 100 are:
\[ 5, 15, 25, 35, 45, 55, 65, 75, 85, 95 \]
These also form an AP where:
- First term \( a = 5 \)
- Common difference \( d = 10 \)
- Last term \( l = 95 \)

Using the same formula to find the number of terms:
\[ n = \frac{95 - 5}{10} + 1 = \frac{90}{10} + 1 = 9 + 1 = 10 \]

### Step 3: Count higher powers of 5 in the first part
Next, we need to count how many of these multiples of 5 are also multiples of \( 25 \) (i.e., \( 5^2 \)):
The odd multiples of 25 less than 100 are:
\[ 25, 75 \]
This is also an AP where:
- First term \( a = 25 \)
- Common difference \( d = 50 \)
- Last term \( l = 75 \)

Calculating the number of terms:
\[ n = \frac{75 - 25}{50} + 1 = \frac{50}{50} + 1 = 1 + 1 = 2 \]

Finally, we check for multiples of \( 125 \) (i.e., \( 5^3 \)):
There are no odd multiples of 125 less than 100.

### Step 4: Total contribution from the first part
The total contribution of the first part to the power of 5 in the product is:
\[ 10 \text{ (from } 5) + 2 \text{ (from } 25) + 0 \text{ (from } 125) = 12 \]

### Step 5: Analyze the second part of the set
The second part consists of even integers greater than 100 and up to 200:
\[ 102, 104, 106, \ldots, 200 \]

This is an AP where:
- First term \( a = 102 \)
- Common difference \( d = 2 \)
- Last term \( l = 200 \)

Finding the number of terms:
\[ n = \frac{200 - 102}{2} + 1 = \frac{98}{2} + 1 = 49 + 1 = 50 \]

### Step 6: Count multiples of 5 in the second part
The even multiples of 5 in this range are:
\[ 110, 120, 130, 140, 150, 160, 170, 180, 190, 200 \]
This is an AP where:
- First term \( a = 110 \)
- Common difference \( d = 10 \)
- Last term \( l = 200 \)

Calculating the number of terms:
\[ n = \frac{200 - 110}{10} + 1 = \frac{90}{10} + 1 = 9 + 1 = 10 \]

### Step 7: Count higher powers of 5 in the second part
Next, we count the multiples of \( 25 \):
The even multiples of 25 in this range are:
\[ 110, 130, 150, 170, 190 \]
This is an AP where:
- First term \( a = 110 \)
- Common difference \( d = 50 \)
- Last term \( l = 190 \)

Calculating the number of terms:
\[ n = \frac{190 - 110}{50} + 1 = \frac{80}{50} + 1 = 1 + 1 = 2 \]

Finally, we check for multiples of \( 125 \):
There are no even multiples of 125 in this range.

### Step 8: Total contribution from the second part
The total contribution of the second part to the power of 5 in the product is:
\[ 10 \text{ (from } 5) + 2 \text{ (from } 25) + 0 \text{ (from } 125) = 12 \]

### Step 9: Combine contributions from both parts
Now, we combine the contributions from both parts:
\[ 12 \text{ (from first part)} + 12 \text{ (from second part)} = 24 \]

### Final Answer
The highest power of 5 that can exactly divide the product of the set \( S \) is \( \boxed{24} \).