Let A = n ∈ N: n is a 3-digit number
B = 9k + 2: k ∈ N
and C = 9k + l: k ∈ N for some l ( 0 < l < 9)
If the sum of all the elements of the set A ∩ (B ∪ C) is 274 × 400, then l is equal to ________.

Solution

Maths sir · 4 steps

Both sets are just 100-term APs of three-digit numbers, so add the two AP sums and solve for l.

1. Three-digit part of B. Three-digit numbers of the form 9k + 2 run from 101 to 992 with common difference 9. That is (992 - 101)/9 + 1 = 100 terms, with sum 100/2(101 + 992) = 54650.
2. Three-digit part of C. For 0 < l < 9, the three-digit numbers of the form 9k + l run from 99 + l to 990 + l. That is again 100 terms, with sum 50(1089 + 2l) = 54450 + 100l.
3. Add the two sums. If l = 2, then C equals B and the total would be only 54650, so l ≠ 2. Then B and C share no element, because they leave different remainders on division by 9. So the sum over the union is the sum of the two parts.

   54650 + 54450 + 100l = 274 × 400 = 109600

4. So. 109100 + 100l = 109600 gives 100l = 500, so l = 5.

The idea to keep. Numbers with a fixed remainder mod 9 form an AP. Find the first and last term in the range, count the terms, and use n/2(a + l). Different remainders mean the sets do not overlap.

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