Let $A = \{n \in \mathbb{N} : \text{H.C.F.}(n, 45) = 1\}$ and $B = \{2k : k \in \{1, 2, \ldots, 100\}\}$. Then the sum of all the elements of $A \cap B$ is ____.

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Sets Relations and Functions

Let

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

and

B = {2 k : k \in {1, 2, \dots, 100}}

. Then the sum of all the elements of

A \cap B

is ____.

Answer: 5264

The set

B

contains all even numbers from 2 to 200. The set

A

contains natural numbers coprime to 45. Since

45 = 3^{2} \times 5

, a number is coprime to 45 if and only if it is not divisible by 3 or 5.

For

A \cap B

, we need even numbers from 2 to 200 that are coprime to 45 (not divisible by 3 or 5).

Using inclusion-exclusion:

Sum = multiples of 2 \sum - multiples of 6 \sum - multiples of 10 \sum + multiples of 30 \sum

= (2 + 4 + \dots + 200) - (6 + 12 + \dots + 198) - (10 + 20 + \dots + 200) + (30 + 60 + \dots + 180)

= \frac{100 \times 202}{2} - \frac{33 \times 204}{2} - \frac{20 \times 210}{2} + \frac{6 \times 210}{2}

= 10100 - 3366 - 2100 + 630 = 5264

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Integer answer