JEE MAIN - Mathematics (2022 - 26th June Morning Shift - No. 18)

Let A = {n $$\in$$ N : H.C.F. (n, 45) = 1} and

Let B = {2k : k $$\in$$ {1, 2, ......., 100}}. Then the sum of all the elements of A $$\cap$$ B is ____________.

Answer
5264

Explanation

Sum of all elements of A $$\cap$$ B = 2 [Sum of natural numbers upto 100 which are neither divisible by 3 nor by 5]

$$ = 2\left[ {{{100 \times 101} \over 2} - 3\left( {{{33 \times 34} \over 2}} \right) - 5\left( {{{20 \times 21} \over 2}} \right) + 15\left( {{{6 \times 7} \over 2}} \right)} \right]$$

$$ = 10100 - 3366 - 2100 + 630$$

$$ = 5264$$

Comments (0)

Advertisement