MCQ
The sum of all the elements of the set $\{\alpha \in\{1,2, \ldots, 100\}: \operatorname{HCF}(\alpha, 24)=1\}$ is
- A$1485$
- ✓$1633$
- C$1857$
- D$1578$
Now, $24=2^{2} \cdot 3$
$\rightarrow \alpha$ is not the multiple of $2$ or $3$
Sum of values of $\alpha$
$= S ( U )-\{ S$ (multiple of $2$)$+ S$ (multiple of$3$ )
- $S$ (multiple of $6$)
$=(1+2+3+\ldots . .100)-(2+4+6 \ldots .+100)-(3$
$+6+\ldots . .99)+(6+12+\ldots .+96)$
$=\frac{100 \times 101}{2}-50 \times 51-\frac{33}{2} \times(3+99)+\frac{16}{2}(6+96)$
$=5050-2550-1683+816=1633$
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