JEE Main 2022 — Sets and Relations Question with Solution

Given $S=\left\{1,2,3\dots 100\right\}$

Now finding the sum of $S=\frac{100\times 101}{2}$

 Prime factors of $24=2^3\times 3$

Let $n\left(A\right)=$ Multiples of $2$

$n\left(B\right)=$ Multiples of $3$

$n\left(A\cap B\right)=$ Multiples of $2$ & $3$

So $n\left(A\cup B\right)=n\left(A\right)+n\left(B\right)-n\left(A\cap B\right)$

To have H.C.F to be $1$ we need to subtract the sum of multiples of $2 \& 3$ from sum of set $S$ to get required answer,

So required answer

$=\frac{100\times 101}{2}-$ Sum of $n\left(A\cup B\right)$

$=\frac{100\times 101}{2}-\left\{2\times \frac{50\times 51}{2}+\frac{33}{2}\left(102\right)-\frac{16}{2}\times 102\right\}$ $=1633$