JEE Mains · Maths · STD 11 - 8. sequence and series
The sum of all the elements of the set \(\{\alpha \in\{1,2, \ldots, 100\}: \operatorname{HCF}(\alpha, 24)=1\}\) is
- A \(1485\)
- B \(1633\)
- C \(1857\)
- D \(1578\)
Answer & Solution
Correct Answer
(B) \(1633\)
Step-by-step Solution
Detailed explanation
\(\operatorname{HCF}(\alpha, 24)=1\) 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 \(6\))…
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