Question

Let A = {n∈N : H.C.F. (n, 45) = 1} and
Let B = {2k :k∈ {1, 2, …,100}}. Then the sum of all the elements of $A∩B$ is ___________

Answer 1

The correct answer is 5264
Sum of all elements of $A∩B$ = 2 [Sum of natural numbers upto 100 which are neither divisible by 3 nor by 5]
$=2[\frac{100×101}{2}−3(\frac{33×34}{2})−5(\frac{20×21}{2})+15(\frac{6×7}{2})]$
= 10100 – 3366 – 2100 + 630
= 5264