2016 · Q28
Let S be a set of first fourteen natural numbers. The possible number of pairs (a, b), where a, b ∈ S and a≠ b such that ab leaves remainder 1 when divided by 15, is
Elementary MathematicsSet Theory
- A3
- B5
- C6
- DNone of the above
Explanation
We need ab ≡ 1 (mod 15) with a ≠ b from {1,…,14}. The invertible residues pair up as (2,8), (8,2), (7,13), (13,7); the self-inverse ones 1, 4, 11, 14 are excluded by a ≠ b. That gives 4 ordered (2 unordered) pairs — none of 3, 5 or 6, so 'None of the above', option D.