Let A = n ∈ N | n² ≤ n + 10,000, B = 3k + 1 | k∈ N an dC = 2k | k∈N, then the sum of all the elements of the set A ∩(B - C) is equal to _____________.

Solution

Maths sir · 4 steps

Pin down A as a plain range first, then keep only the odd members of B inside it.

1. Find A. The condition is n(n-1) ≤ 10000. For n = 100, 100 × 99 = 9900 ≤ 10000. For n = 101, 101 × 100 = 10100 > 10000. So A = {1, 2, …, 100}.
2. Find B minus C. B = {4, 7, 10, 13, …} because k starts at 1. C is the set of even natural numbers, so removing it keeps the odd members of B. These are 7, 13, 19, …, an AP with common difference 6.
3. Intersect with A. Keep the terms up to 100: 7, 13, …, 97. The number of terms is (97 - 7)/6 + 1 = 16.

   S = 16/2(7 + 97) = 8 × 104 = 832

4. So. The sum of all elements of A ∩ (B - C) is 832.

The common slip: 1716. Forgetting to remove C and adding every member of B from 4 to 100 gives 1716. The 17 even terms, which add up to 884, must be taken out.

The idea to keep. Turn each set into a list or a pattern first. Removing the evens from an AP with odd common difference leaves every second term, which is a new AP you can sum directly.

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