Mathematics - Permutation & Combination Question with Solution | TestHub

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

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

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

$=1251$

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

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

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

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

$=2500-867-365-17$

$=1251.$