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 ____________.

Solution

Maths sir · 3 steps

Start with all even numbers up to 200, then remove multiples of 3 and of 5 using inclusion–exclusion.

1. Read the sets. Since 45 = 3² × 5, set A holds the natural numbers divisible by neither 3 nor 5. Set B is the even numbers 2, 4, …, 200, so A ∩ B is the even numbers up to 200 that are not multiples of 3 or 5.

   Even numbers that are multiples of	Terms	Sum
   2 (all of B)	2, 4, …, 200: 100 terms	2 × (100 × 101)/2 = 10100
   6 (even and divisible by 3)	6, 12, …, 198: 33 terms	6 × (33 × 34)/2 = 3366
   10 (even and divisible by 5)	10, 20, …, 200: 20 terms	10 × (20 × 21)/2 = 2100
   30 (even and divisible by both)	30, 60, …, 180: 6 terms	30 × (6 × 7)/2 = 630

2. Inclusion–exclusion. Multiples of 30 get removed twice, once with the 6s and once with the 10s, so add them back once.

   10100 - 3366 - 2100 + 630 = 5264

3. So. The sum of all elements of A ∩ B is 5264.

The common slip: 4634. Forgetting to add back the multiples of 30 gives 10100 - 3366 - 2100 = 4634, not 5264.

The idea to keep. For a coprimality condition, factorise the number first. Then sum what is allowed with inclusion–exclusion: subtract each prime's multiples and add back the overlaps.

Answer from the NTA final answer key. Solution by PrepWiser.