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

We need the sum of all $$\alpha \in \{1, 2, \ldots, 100\}$$ such that $$\gcd(\alpha, 24) = 1$$.

$$24 = 2^3 \times 3$$. So $$\gcd(\alpha, 24) = 1$$ means $$\alpha$$ is not divisible by 2 or 3.

We use inclusion-exclusion. The sum of all integers from 1 to 100 is $$\frac{100 \times 101}{2} = 5050$$.

Sum of multiples of 2 from 1 to 100: $$2 + 4 + \ldots + 100 = 2 \times \frac{50 \times 51}{2} = 2550$$.

Sum of multiples of 3 from 1 to 100: $$3 + 6 + \ldots + 99 = 3 \times \frac{33 \times 34}{2} = 3 \times 561 = 1683$$.

Sum of multiples of 6 from 1 to 100: $$6 + 12 + \ldots + 96 = 6 \times \frac{16 \times 17}{2} = 6 \times 136 = 816$$.

By inclusion-exclusion, the sum of integers divisible by 2 or 3:

$$S_{2 \cup 3} = 2550 + 1683 - 816 = 3417$$

The sum of integers coprime to 24:

$$S = 5050 - 3417 = 1633$$

The correct answer is $$1633$$.