The sum of all the elements of the set `{alpha in {1,2,…….,100}" HCF " (alpha,24) =1}` is

Answer

To solve the problem of finding the sum of all elements of the set `{alpha in {1,2,…….,100} | HCF(alpha, 24) = 1}`, we need to follow these steps:

### Step 1: Identify the elements of the set
We need to find all integers \( \alpha \) from 1 to 100 such that \( \text{HCF}(\alpha, 24) = 1 \). This means that \( \alpha \) must not share any prime factors with 24.

### Step 2: Factorize 24
The prime factorization of 24 is:
\[ 24 = 2^3 \times 3^1 \]
Thus, the prime factors of 24 are 2 and 3.

### Step 3: Identify numbers coprime to 24
A number is coprime to 24 if it is not divisible by either 2 or 3. Therefore, we need to find all numbers from 1 to 100 that are not divisible by 2 or 3.

### Step 4: Count the numbers divisible by 2 or 3
We can use the principle of inclusion-exclusion to count the numbers divisible by 2 or 3.

1. Count of numbers divisible by 2:
\[ \text{Count}_{2} = \left\lfloor \frac{100}{2} \right\rfloor = 50 \]

2. Count of numbers divisible by 3:
\[ \text{Count}_{3} = \left\lfloor \frac{100}{3} \right\rfloor = 33 \]

3. Count of numbers divisible by both 2 and 3 (i.e., divisible by 6):
\[ \text{Count}_{6} = \left\lfloor \frac{100}{6} \right\rfloor = 16 \]

Using inclusion-exclusion, the total count of numbers divisible by either 2 or 3 is:
\[ \text{Count}_{2 \cup 3} = \text{Count}_{2} + \text{Count}_{3} - \text{Count}_{6} = 50 + 33 - 16 = 67 \]

### Step 5: Count numbers coprime to 24
The total numbers from 1 to 100 is 100. Therefore, the count of numbers that are coprime to 24 is:
\[ \text{Count}_{\text{coprime}} = 100 - \text{Count}_{2 \cup 3} = 100 - 67 = 33 \]

### Step 6: List the numbers coprime to 24
The numbers coprime to 24 from 1 to 100 are:
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.

### Step 7: Calculate the sum of these numbers
Now, we sum these numbers:
\[ \text{Sum} = 1 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 25 + 29 + 31 + 35 + 37 + 41 + 43 + 47 + 49 + 53 + 55 + 59 + 61 + 65 + 67 + 71 + 73 + 77 + 79 + 83 + 85 + 89 + 91 + 95 + 97 \]

Calculating this gives:
\[ \text{Sum} = 1956 \]

### Final Answer
Thus, the sum of all elements of the set is:
\[ \boxed{1956} \]