The sum of all the elements of the set \ α ∈ \ 1, 2, . . 100\ :...

Given $S = \left\{1, 2, 3 \ldots 100\right\}$

Now finding the sum of $S = \dfrac{100 \times 101}{2}$

Prime factors of $24 = {2}^{3} \times 3$

Let $n\left(A\right) =$ Multiples of $2$

$n\left(B\right) =$ Multiples of $3$

$n\left(A \cap B\right) =$ Multiples of $2$ & $3$

So $n\left(A \cup B\right) = n\left(A\right) + n\left(B\right) - n\left(A \cap B\right)$

To have H.C.F to be $1$ we need to subtract the sum of multiples of $2 \text{ and } 3$ from sum of set $S$ to get required answer,

So required answer

$= \dfrac{100 \times 101}{2} -$ Sum of $n\left(A \cup B\right)$

$= \dfrac{100 \times 101}{2} - \left\{2 \times \dfrac{50 \times 51}{2} + \dfrac{33}{2}\left(102\right) - \dfrac{16}{2} \times 102\right\}$ $= 1633$