The sum of all the elements of the set {alpha | vedclass

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Question

$\operatorname{HCF}(\alpha, 24)=1$

Now, $24=2^{2} \cdot 3$

$\rightarrow \alpha$ is not the multiple of $2$ or $3$

Sum of values of $\alpha$

Vedclass · Accepted Answer

$1633$

Vedclass · Answer

$\operatorname{HCF}(\alpha, 24)=1$

Now, $24=2^{2} \cdot 3$

$ightarrow \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 $6$)

$=(1+2+3+\ldots . .100)-(2+4+6 \ldots .+100)-(3$

$+6+\ldots . .99)+(6+12+\ldots .+96)$

$=\frac{100 	imes 101}{2}-50 	imes 51-\frac{33}{2} 	imes(3+99)+\frac{16}{2}(6+96)$

$=5050-2550-1683+816=1633$

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

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