Sum of all elements of set {α ∈{1,2, ldots, 100}: operatorname{HCF}(α, 24)=1} is

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8. Sequences and Series

hard

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

$1485$

$1633$

$1857$

$1578$

(JEE MAIN-2022)

$\operatorname{HCF}(\alpha, 24)=1$

Now, $24=2^{2} \cdot 3$

$\rightarrow \alpha$ is not the multiple of $2$ or $3$

Sum of values of $\alpha$

$= S ( U )-\{ S$ (multiple of $2$)$+ S$ (multiple of$3$ )

– $S$ (multiple of $6$)

$=(1+2+3+\ldots . .100)-(2+4+6 \ldots .+100)-(3$

$+6+\ldots . .99)+(6+12+\ldots .+96)$

$=\frac{100 \times 101}{2}-50 \times 51-\frac{33}{2} \times(3+99)+\frac{16}{2}(6+96)$

$=5050-2550-1683+816=1633$

Standard 11

Mathematics

easy

easy

easy

hard

(KVPY-2021)

medium