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 odd integers less than 100
The odd integers less than 100 are:
\[ 1, 3, 5, 7, 9, \ldots, 99 \]

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

To find the number of terms \( n \):
\[ n = \frac{l - a}{d} + 1 = \frac{99 - 1}{2} + 1 = 50 \]

Next, we need to find how many of these terms are divisible by 5:
The odd multiples of 5 less than 100 are:
\[ 5, 15, 25, 35, 45, 55, 65, 75, 85, 95 \]

Counting these gives us 10 terms. Now, we need to determine how many times 5 divides each of these terms:
- \( 5^1 \) from \( 5 \)
- \( 5^1 \) from \( 15 \)
- \( 5^2 \) from \( 25 \)
- \( 5^1 \) from \( 35 \)
- \( 5^1 \) from \( 45 \)
- \( 5^1 \) from \( 55 \)
- \( 5^1 \) from \( 65 \)
- \( 5^1 \) from \( 75 \)
- \( 5^1 \) from \( 85 \)
- \( 5^1 \) from \( 95 \)

Calculating the total contributions:
- From \( 5 \): 1
- From \( 15 \): 1
- From \( 25 \): 2
- From \( 35 \): 1
- From \( 45 \): 1
- From \( 55 \): 1
- From \( 65 \): 1
- From \( 75 \): 1
- From \( 85 \): 1
- From \( 95 \): 1

Total power of 5 from odd integers:
\[ 1 + 1 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12 \]

### Step 2: Analyze the even integers from 102 to 200
The even integers greater than 100 and up to 200 are:
\[ 102, 104, 106, \ldots, 200 \]

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

To find the number of terms \( n \):
\[ n = \frac{l - a}{d} + 1 = \frac{200 - 102}{2} + 1 = 50 \]

Now, we need to find how many of these terms are divisible by 5:
The even multiples of 5 in this range are:
\[ 110, 120, 130, 140, 150, 160, 170, 180, 190, 200 \]

Counting these gives us 10 terms. Now, we need to determine how many times 5 divides each of these terms:
- \( 5^1 \) from \( 110 \)
- \( 5^2 \) from \( 120 \)
- \( 5^1 \) from \( 130 \)
- \( 5^2 \) from \( 140 \)
- \( 5^2 \) from \( 150 \)
- \( 5^1 \) from \( 160 \)
- \( 5^1 \) from \( 170 \)
- \( 5^1 \) from \( 180 \)
- \( 5^1 \) from \( 190 \)
- \( 5^2 \) from \( 200 \)

Calculating the total contributions:
- From \( 110 \): 1
- From \( 120 \): 2
- From \( 130 \): 1
- From \( 140 \): 2
- From \( 150 \): 2
- From \( 160 \): 1
- From \( 170 \): 1
- From \( 180 \): 1
- From \( 190 \): 1
- From \( 200 \): 2

Total power of 5 from even integers:
\[ 1 + 2 + 1 + 2 + 2 + 1 + 1 + 1 + 1 + 2 = 12 \]

### Step 3: Combine the results
The total power of 5 that divides the product of all numbers in set \( S \) is:
\[ 12 \text{ (from odd integers)} + 12 \text{ (from even integers)} = 24 \]

### Final Answer
The highest power of 5 that can exactly divide the product is:
\[ \boxed{24} \]