Let S be a set of first 10 natural numbers. What is the possible number of pairs (a, b) where a, b ∈ S and a ≠ b such that the product ab (>12) leaves remainder 4 when divided by 12?

S is a set of first 10 natural numbers;

∴ S = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

∵ a, b ∈ S and a ≠ b such that ab > 12;

∴ The product ab can’t be any 3 digit number (it will be 2 digit number)

∴ Product ab (>12) that leaves remainder 4 when divided by 12 = (12n + 4);

∴ All such number (2 digit only) will be 16, 28, 40, 52, 64, 76, 88

Now classifying these numbers as the product ab in the pair (a, b) where a, b ∈ S a ≠ b:

16: (2, 8), (8, 2)

28: (4, 7), (7, 4)

40: (5, 8), (8, 5), (4, 10), (10, 4)

52: None

64: None

76: None

88: None

∴ Number of total pairs = 8