Key Concepts and Formulas

• Coprime Numbers: Two integers are coprime (or relatively prime) if their greatest common divisor (GCD) is 1. The condition $HCF(\alpha, n) = 1$ means $\alpha$ is coprime to $n$.
• Prime Factorization: To determine coprimality with $n$, we need the prime factorization of $n$. A number $\alpha$ is coprime to $n$ if it shares no prime factors with $n$.
• Euler's Totient Function ($\phi(n)$): Counts the number of positive integers up to $n$ that are relatively prime to $n$. If $n = p_1^{k_1} p_2^{k_2} \dots p_r^{k_r}$, then $\phi(n) = n \prod_{i=1}^{r} (1 - \frac{1}{p_i})$.
• Periodicity of Coprimality: The property of being coprime to $n$ is periodic with period $n$. If $HCF(\alpha, n) = 1$, then $HCF(\alpha + kn, n) = 1$ for any integer $k$.
• Sum of Coprime Numbers in $[1, n]$ (for $n>2$): The sum of positive integers less than or equal to $n$ that are coprime to $n$ is $\frac{n \cdot \phi(n)}{2}$.

Step-by-Step Solution

Step 1: Understand the Problem and Prime Factorize 24 We need to find the sum of all numbers $\alpha$ in the set $\{1, 2, \ldots, 100\}$ such that $HCF(\alpha, 24) = 1$. This means $\alpha$ must not share any prime factors with 24. The prime factorization of 24 is $24 = 2^3 \times 3^1$. Therefore, for $HCF(\alpha, 24) = 1$, $\alpha$ must not be divisible by 2 and must not be divisible by 3.

Step 2: Analyze the Coprime Numbers in the First Block [1, 24] First, we determine how many numbers between 1 and 24 are coprime to 24. This is given by Euler's totient function $\phi(24)$. $\phi(24) = 24 \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3}\right) = 24 \left(\frac{1}{2}\right) \left(\frac{2}{3}\right) = 8$ There are 8 numbers in the range $[1, 24]$ that are coprime to 24. These numbers are those not divisible by 2 or 3: $\{1, 5, 7, 11, 13, 17, 19, 23\}$. Let $S_0$ be the sum of these numbers: $S_0 = 1 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 96$ Alternatively, using the formula for the sum of coprime numbers up to $n$ (for $n>2$): $S_0 = \frac{24 \cdot \phi(24)}{2} = \frac{24 \cdot 8}{2} = 96$

Step 3: Utilize Periodicity to Sum Over Full Blocks The range of numbers is $\{1, 2, \ldots, 100\}$. We can divide this range into blocks of 24. $100 = 4 \times 24 + 4$. This means we have 4 full blocks of 24 numbers and a partial block of 4 numbers. The full blocks cover the range $[1, 96]$. The blocks are:

• Block 0: $[1, 24]$
• Block 1: $[25, 48]$
• Block 2: $[49, 72]$
• Block 3: $[73, 96]$

The property of coprimality is periodic with period 24. If $x$ is coprime to 24, then $x + 24k$ is also coprime to 24. Let $S_k$ be the sum of numbers coprime to 24 in the $k$-th block, which spans from $24k+1$ to $24(k+1)$. The numbers coprime to 24 in this block are of the form $24k + x$, where $x \in \{1, 5, 7, 11, 13, 17, 19, 23\}$. The sum for the $k$-th block is: $S_k = \sum_{x \in \{1,5,...,23\}} (24k + x) = \phi(24) \times 24k + \sum_{x \in \{1,5,...,23\}} x$ $S_k = 8 \times 24k + S_0 = 192k + 96$

We need to sum $S_k$ for $k=0, 1, 2, 3$ (for the full blocks): $\text{Sum}_{\text{full blocks}} = \sum_{k=0}^{3} S_k = \sum_{k=0}^{3} (192k + 96)$ $= 192 \sum_{k=0}^{3} k + \sum_{k=0}^{3} 96$ $= 192 (0+1+2+3) + 4 \times 96$ $= 192 \times 6 + 384$ $= 1152 + 384 = 1536$

Step 4: Sum Numbers in the Partial Block [97, 100] The remaining numbers are in the range $[97, 100]$. We check each of these numbers for coprimality with 24 (i.e., not divisible by 2 or 3).

• 97: 97 is not divisible by 2 (it's odd) and not divisible by 3 (sum of digits $9+7=16$, not divisible by 3). So, $HCF(97, 24) = 1$.
• 98: 98 is divisible by 2. So, $HCF(98, 24) \ne 1$.
• 99: 99 is divisible by 3 (sum of digits $9+9=18$, divisible by 3). So, $HCF(99, 24) \ne 1$.
• 100: 100 is divisible by 2. So, $HCF(100, 24) \ne 1$.

Only 97 is coprime to 24 in this partial block. The sum for the partial block is 97.

Step 5: Calculate the Total Sum The total sum is the sum from the full blocks plus the sum from the partial block. $\text{Total Sum} = \text{Sum}_{\text{full blocks}} + \text{Sum}_{\text{partial block}}$ $\text{Total Sum} = 1536 + 97 = 1633$

Common Mistakes & Tips

• Incorrect Prime Factorization: Ensure the prime factorization of 24 is correct. Errors here will lead to incorrect coprimality checks.
• Forgetting the Partial Block: Always check the numbers in the remaining partial range after accounting for full blocks.
• Arithmetic Errors: Summing over multiple blocks can lead to calculation mistakes. Double-check your arithmetic.
• Misunderstanding Coprimality: Remember that $HCF(\alpha, 24)=1$ means $\alpha$ is not divisible by 2 AND not divisible by 3.

Summary To find the sum of elements $\alpha$ in $\{1, 2, \ldots, 100\}$ such that $HCF(\alpha, 24) = 1$, we first identified that $\alpha$ must not be divisible by 2 or 3. We then used Euler's totient function to find that there are 8 numbers coprime to 24 in any block of 24. The sum of these numbers in the first block $[1, 24]$ is 96. Leveraging the periodicity of coprimality, we calculated the sum of coprime numbers in the four full blocks from 1 to 96, which is 1536. Finally, we checked the numbers in the partial block $[97, 100]$ and found that only 97 is coprime to 24. Adding this to the sum of the full blocks gives the total sum.

The final answer is $\boxed{1633}$.