If S is given as `{S: 1,2,3,4, . . . 100}` then find the sum of value of S such that `"{HCF of 24 and S is 1}"`

Answer

To solve the problem of finding the sum of values in the set \( S = \{1, 2, 3, \ldots, 100\} \) such that the highest common factor (HCF) of 24 and \( S \) is 1, we can follow these steps:

### Step 1: Understand the condition for HCF
The condition \( \text{HCF}(24, S) = 1 \) means that the numbers in \( S \) must not share any prime factors with 24.

### Step 2: Factorize 24
The prime factorization of 24 is:
\[ 24 = 2^3 \times 3^1 \]
This means that any number \( s \in S \) must not be divisible by 2 or 3.

### Step 3: Identify the numbers in \( S \)
The set \( S \) consists of the integers from 1 to 100:
\[ S = \{1, 2, 3, 4, \ldots, 100\} \]

### Step 4: Count numbers not divisible by 2 or 3
To find the numbers that are not divisible by 2 or 3, we can use the principle of inclusion-exclusion.

1. **Count multiples of 2 in \( S \)**:
The multiples of 2 from 1 to 100 are:
\[ 2, 4, 6, \ldots, 100 \]
This is an arithmetic series where:
- First term \( a = 2 \)
- Common difference \( d = 2 \)
- Last term \( l = 100 \)
The number of terms \( n \) can be found using:
\[ n = \frac{l - a}{d} + 1 = \frac{100 - 2}{2} + 1 = 50 \]

2. **Count multiples of 3 in \( S \)**:
The multiples of 3 from 1 to 100 are:
\[ 3, 6, 9, \ldots, 99 \]
Similarly, this is an arithmetic series where:
- First term \( a = 3 \)
- Common difference \( d = 3 \)
- Last term \( l = 99 \)
The number of terms \( n \) is:
\[ n = \frac{l - a}{d} + 1 = \frac{99 - 3}{3} + 1 = 33 \]

3. **Count multiples of 6 in \( S \)** (to avoid double counting):
The multiples of 6 from 1 to 100 are:
\[ 6, 12, 18, \ldots, 96 \]
This is again an arithmetic series where:
- First term \( a = 6 \)
- Common difference \( d = 6 \)
- Last term \( l = 96 \)
The number of terms \( n \) is:
\[ n = \frac{l - a}{d} + 1 = \frac{96 - 6}{6} + 1 = 16 \]

### Step 5: Apply inclusion-exclusion principle
Using the inclusion-exclusion principle, the total number of integers from 1 to 100 that are divisible by either 2 or 3 is:
\[ \text{Total} = (\text{Multiples of 2}) + (\text{Multiples of 3}) - (\text{Multiples of 6}) = 50 + 33 - 16 = 67 \]

### Step 6: Calculate the count of numbers not divisible by 2 or 3
The total count of integers from 1 to 100 is 100. Thus, the count of integers not divisible by either 2 or 3 is:
\[ 100 - 67 = 33 \]

### Step 7: Calculate the sum of numbers not divisible by 2 or 3
The numbers not divisible by 2 or 3 in the range from 1 to 100 can be listed as:
\[ 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101 \]
To find the sum of these numbers, we can use the formula for the sum of an arithmetic series or simply add them up.

### Final Calculation
The sum of these numbers can be calculated directly or using a programming tool or calculator. After calculating, we find:
\[ \text{Sum} = 1633 \]

### Final Answer
The sum of the values of \( S \) such that \( \text{HCF}(24, S) = 1 \) is:
\[ \boxed{1633} \]