JEE Mains · Maths · STD 11 - 8. sequence and series
The sum of all the elements in the set \(\{\mathrm{n} \in\{1,2, \ldots \ldots ., 100\} \mid\) \(H.C.F.\) of \(n\) and \(2040\) is \(1\,\}\) is equal to \(.....\)
- A \(1251\)
- B \(1300\)
- C \(1456\)
- D \(1371\)
Answer & Solution
Correct Answer
(A) \(1251\)
Step-by-step Solution
Detailed explanation
\(2040=2^{3} \times 3 \times 5 \times 17\) \(n\) should not be multiple of \(2,3,5\) and \(17 .\) Sum of all \(n=(1+3+5 \ldots . .+99)-(3+9+15+21+\ldots . .+99)-(5+25+35+55+65\) \(+85+95)-(17)\) \(=2500-\frac{17}{2}(3+99)-365-17\) \(=2500-867-365-17=1251\)
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