Question

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

Answer 1

$\because 2040=2^3\cdot 3\cdot 5\cdot 17$
Let $A=$ Sum of all numbers which are divisible by $2$ upto $100.$
$B=$ Sum of all numbers which are divisilbe by $3$ upto $100$
$C=$ Sum of all numbers which are divisible by $5$ upto $100$
$D=$ Sum of all numbers which are divisilbe by $17$ upto $100$
$A\cup B\cup C\cup D=(A+B+C+D)-(A\cap B+A\cap C+A\cap D+B\cap C+B\cap D+C\cap D)+(A\cap B\cap C+A\cap B\cap D+A\cap C\cap D+B\cap C\cap D)-(A\cap B\cap C\cap D)$
$=(50\times 51+33\times 51+1050+51\times 5)-(51\times 16+550+102+315+51+85)+(180+0+0+0)-0=3799$
Required sum $=5050-3799=1251$