Let S = n ∈ N , 0 i 1 0 n a b c d = a b c d ∀ a , b , c , d ∈ R , where i = - 1 . Then the number of 2 - digit numbers in the set S is

Step-by-step solution

Let X = a b c d   &   A = 0 i 1 0 n ⇒ A X = I X ⇒ A = I ⇒ 0 i 1 0 n = I ⇒ A 8 = 1 0 0 1 ⇒ n is multiple of 8 . So number of 2 digit numbers in the set S = 16 , 24 , 32 , … , 96 Clearly, 16 , 24 , 32 , … , 96 are in arithmetic progression. Here, a = 16 ,   d = 24 - 16 = 8 and l = a + n - 1 d = 96 Where, a is the first term, d is the common difference, l is the last term and n is the number of terms. Consider, l = a + n - 1 d = 96 ⇒ 16 + n - 1 8 = 96