Question:
The sum of all the elements in the set $\{n \in\{1,2, \ldots \ldots, 100\}$ । H.C.F. of $\mathrm{n}$ and 2040 is 1$\}$ is equal to
Solution:
$2040=2^{3} \times 3 \times 5 \times 17$
$\mathrm{n}$ should not be multiple of $2,3,5$ and 17 .
Sum of all $n=(1+3+5 \ldots \ldots+99)-(3+9+15+$ $21+\ldots \ldots+99)-(5+25+35+55+65+85+95)$ $-(17)$
$=2500-\frac{17}{2}(3+99)-365-17$
$=1251$
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