The sum of all the elements of the set {α ∈ {1, 2, ..., 100} : HCF (α, 24) = 1} is ____.

Q: Wait, why do we subtract $S_6$? Isn't that double counting?

Exactly! When you subtract the multiples of 2, you remove 6. When you subtract the multiples of 3, you remove 6 again. You've removed it twice! Adding $S_6$ back (or rather, subtracting $S_6$ once in the PIE formula $S_2 + S_3 - S_6$) ensures 6 is removed exactly once.

Q: Can't I just list all the numbers and add them up?

You could, but in a JEE exam, time is your most precious resource. Listing numbers is a trap that leads to calculation errors and wasted minutes. The PIE method is elegant, systematic, and virtually error-proof.

Q: What if the question asked for numbers coprime to 100 instead of 24?

The logic remains identical! Find the prime factors of 100 ($2^2 \times 5^2$), so the prime factors are 2 and 5. You would then find the sum of multiples of 2, the sum of multiples of 5, and subtract the sum of multiples of 10. The beauty of this method is its scalability.

Question

The sum of all the elements of the set $\{\alpha \in \{1, 2, \dots, 100\} : 	ext{HCF}(\alpha, 24) = 1\}$ is ____.

Sigma Percentile · Accepted Answer

### Analyzing the Universe of Numbers

Imagine you are standing at the threshold of a vast, orderly universe: the set of integers from 1 to 100. Your task is to find the sum of all elements $\alpha$ that share a special relationship with the number 24—they must be coprime, meaning $	ext{HCF}(\alpha, 24) = 1$.

This is essentially a sieve process. We are filtering out the 'noise' to find the 'signal'.

### Decoding the Condition

To understand what makes a number 'coprime' to 24, we must look at its prime factorization. The prime factorization of 24 is:

$$24 = 2^3 	imes 3$$

This tells us that any number sharing a factor with 24 must be divisible by either 2 or 3. If a number is divisible by 2, it is excluded. If it is divisible by 3, it is also excluded.

Our goal is to find the sum of all numbers in our set that are neither divisible by 2 nor by 3.

### The Strategy: The Principle of Inclusion-Exclusion

Instead of hunting for these 'good' numbers one by one, we use the **Principle of Inclusion-Exclusion (PIE)**. We start with the total sum of all numbers from 1 to 100, denoted as $S_{	ext{total}}$.

From this, we subtract the sum of all multiples of 2 ($S_2$) and the sum of all multiples of 3 ($S_3$). However, when we subtract $S_2$ and $S_3$, we have subtracted the multiples of 6 (the intersection of 2 and 3) twice.

To correct this, we must add the sum of multiples of 6 ($S_6$) back in. The master formula is:

$$S_{	ext{required}} = S_{	ext{total}} - (S_2 + S_3 - S_6)$$

### The Arithmetic Grind

First, the total sum of the first 100 natural numbers is given by the formula $S_n = \frac{n(n+1)}{2}$. For $n=100$:

$$S_{	ext{total}} = \frac{100 	imes 101}{2} = 5050$$

Next, we calculate the sum of multiples of 2 ($S_2$):

$$S_2 = 2 + 4 + \dots + 100 = 2(1 + 2 + \dots + 50) = 2 	imes \frac{50 	imes 51}{2} = 2550$$

Then, we calculate the sum of multiples of 3 ($S_3$):

$$S_3 = 3 + 6 + \dots + 99 = 3(1 + 2 + \dots + 33) = 3 	imes \frac{33 	imes 34}{2} = 1683$$

Finally, we calculate the sum of multiples of 6 ($S_6$):

$$S_6 = 6 + 12 + \dots + 96 = 6(1 + 2 + \dots + 16) = 6 	imes \frac{16 	imes 17}{2} = 816$$

### The Grand Finale

Now, we combine these pieces to find the sum of the 'bad' numbers:

$$S(2 \cup 3) = S_2 + S_3 - S_6 = 2550 + 1683 - 816 = 3417$$

Finally, we subtract this from our total sum to find the result:

$$S_{	ext{required}} = 5050 - 3417 = 1633$$

You have successfully navigated the sieve. The sum of all elements $\alpha$ coprime to 24 is **1633**.

Animated Solution for Mathematics - Sequence and Series: The sum of all the elements of the set ${α \in {1, 2, \dots, 100} : HCF (α, 24) = 1}$ is ____.

Enter Numerical Value:

Visualized Solution

Visualizing the Universe

Prime Factorization of $24$

The Strategy: Inclusion-Exclusion

Step 1: Total Sum ( $S_{total}$ )

Step 2: Sum of Multiples of $2$ ( $S_{2}$ )

Step 3: Sum of Multiples of $3$ ( $S_{3}$ )

Step 4: Sum of Multiples of $6$ ( $S_{6}$ )

Step 5: Sum of Multiples of 2 or 3

Final Calculation

Conclusion & Key Takeaway

The Sigma Insight: Sum of Special Series

Analyzing the Universe of Numbers

Decoding the Condition

The Strategy

The Arithmetic Grind

The Grand Finale

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of all the elements of the set {α∈{1,2,…,100}:HCF(α,24)=1} is ____.

Enter Numerical Value:

Visualized Solution

Visualizing the Universe

Prime Factorization of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 24

The Strategy: Inclusion-Exclusion

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