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

Correct answer: 5264

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

\( = 2 \left[ \left(\frac{100\times 101}2\right) - 3\left(\frac{33 \times 34}{2} \right) - 5 \left( \frac{20 \times 21}2\right) + 15 \left(\frac{6\times7}2\right)\right]\)

\(= 10100 - 3366 - 2100 + 630\)

\(= 10730 -5466\)

\(= 5264\)

Answer 2

Sum of elements in A ∩ B

 = 5264