The sum of all the elements in the set n ∈ 1, 2, , 100 ∣ H.C.F. of...

The prime factorisation of the given number is $2040 = {2}^{3} \times 3 \times 5 \times 17$

The H.C.F. of $n$ and $2040$ is $1$ if $n$ is not a multiple of $2, 3, 5$ and $17 .$

Hence $n$ cannot be multiple of $2, 3, 5$ and $17$

Then sum is $n ( 1 ) - ( n ( 2 ) + n ( 3 ) + n ( 5 ) + n ( 17 ) - n ( 6 ) - n ( 10 ) - n ( 34 ) - n ( 15 ) - n ( 51 ) - n ( 85 ) ) + n ( 30 ) )$
Where $n ( a )$ means the sum of all numbers belonging to the set ${ 1, 2, \ldots \ldots . 100 }$ which are divisible by $a$

$= \dfrac{100 \times 101}{2} - \left(\dfrac{2 \times 50 \times 51}{2} + \dfrac{3 \times 33 \times 34}{2} + \dfrac{5 \times 20 \times 21}{2} + \dfrac{17 \times 5 \times 6}{2} - \dfrac{6 \times 16 \times 17}{2} - \dfrac{10 \times 10 \times 11}{2} - \dfrac{34 \times 2 \times 3}{2} - \dfrac{15 \times 6 \times 7}{2} - 51 - 85 + 180\right)$

$= 5050 - 2550 - 1683 - 1050 - 255 + 816 + 550 + 102 + 315 + 51 + 85 - 180$

$= 1251$

Alternate Solution:
Thus, the sum of all the required numbers taking the common elements only once, is $\left(1 + 3 + 5 + \ldots . . + 99\right) - \left(3 + 9 + 15 + 21 + \ldots \ldots + 99\right) - \left(5 + 25 + 35 + 55 + 65 + 85 + 95\right) - \left(17\right)$

Now, using the sum of $n$ terms of an arithmetic progression i.e. ${S}_{n} = \dfrac{n}{2}\left[a + l\right],$ where $a$ is the first term and $l$ is the last term.

$= \dfrac{50}{2}\left[1 + 99\right] - \dfrac{17}{2}\left[3 + 99\right] - \left(365\right) - 17$

$= \dfrac{50}{2}\left(100\right) - \dfrac{17}{2}\left(102\right) - 365 - 17$

$= 2500 - 867 - 365 - 17$

$= 1251 .$