JEE Main 2022 (Online) 26th June Morning ShiftQuestion 1
Let $A=\{n\in\mathbf{N}:\text{H.C.F.}(n,45)=1\}$ and let $B=\{2k:k\in\{1,2,\ldots,100\}\}$. Then the sum of all the elements of $A\cap B$ is $\_\_\_\_\_\_$ .
Sets, Relations and Function
Answer$5264$
Solution
$A\cap B$: even numbers $2k\le200$ coprime to 45, i.e. not divisible by 3 or 5. Sum of even $n\le200$: $2\cdot\frac{100\cdot101}{2}=10100$. Remove multiples of 6 (up to 198: $6\cdot\frac{33\cdot34}{2}=3366$) and of 10 (up to 200: $10\cdot\frac{20\cdot21}{2}=2100$); add back multiples of 30 (up to 180: $30\cdot21=630$). $10100-3366-2100+630=5264$.