The sum of all the elements in the set $$\{n \in \{1, 2, \ldots, 100\} | \text{H.C.F. of } n \text{ and } 2040 \text{ is } 1\}$$ is equal to ___.

We need the sum of all $$n \in \{1, 2, \ldots, 100\}$$ such that $$\gcd(n, 2040) = 1$$. First, factorize $$2040 = 2^3 \times 3 \times 5 \times 17$$. So $$n$$ must not be divisible by 2, 3, 5, or 17.

We use inclusion-exclusion on the sum. Let $$S_d$$ denote the sum of all multiples of $$d$$ in $$\{1, \ldots, 100\}$$. If there are $$k = \lfloor 100/d \rfloor$$ such multiples, then $$S_d = d \cdot \frac{k(k+1)}{2}$$.

The sum we want is $$S_{\text{total}} - (S_2 + S_3 + S_5 + S_{17}) + (S_6 + S_{10} + S_{34} + S_{15} + S_{51} + S_{85}) - (S_{30} + S_{102} + S_{170} + S_{255}) + S_{510}$$, where terms with $$d > 100$$ contribute zero.

Computing each: the total sum is $$\frac{100 \times 101}{2} = 5050$$. For $$S_2$$: $$k = 50$$, $$S_2 = 2 \times 1275 = 2550$$. For $$S_3$$: $$k = 33$$, $$S_3 = 3 \times 561 = 1683$$. For $$S_5$$: $$k = 20$$, $$S_5 = 5 \times 210 = 1050$$. For $$S_{17}$$: $$k = 5$$, $$S_{17} = 17 \times 15 = 255$$.

For the pairwise terms: $$S_6$$: $$k = 16$$, $$S_6 = 6 \times 136 = 816$$. $$S_{10}$$: $$k = 10$$, $$S_{10} = 10 \times 55 = 550$$. $$S_{34}$$: $$k = 2$$, $$S_{34} = 34 \times 3 = 102$$. $$S_{15}$$: $$k = 6$$, $$S_{15} = 15 \times 21 = 315$$. $$S_{51}$$: $$k = 1$$, $$S_{51} = 51$$. $$S_{85}$$: $$k = 1$$, $$S_{85} = 85$$.

For the triple terms: $$S_{30}$$: $$k = 3$$, $$S_{30} = 30 \times 6 = 180$$. $$S_{102}$$, $$S_{170}$$, $$S_{255}$$ all have $$k = 0$$, contributing nothing. The quadruple term $$S_{510} = 0$$ as well.

Putting it all together: $$5050 - (2550 + 1683 + 1050 + 255) + (816 + 550 + 102 + 315 + 51 + 85) - 180 = 5050 - 5538 + 1919 - 180 = 1251$$.