Let A = 1 , 2 , 3 , 5 , 8 , 9 . Then the number of possible functions f : A → A such that f m · n = f m · f n for every m , n ∈ A with m · n ∈ A is equal to

Step-by-step solution

Given, f m · n = f m · f n Taking m = 1 ⇒ f 1 · n = f 1 · f n ⇒ f 1 = 1 Now taking m = 3   &   n = 3 we get, f ( 9 ) = f ( 3 ) × f ( 3 ) , So, f 3 ,   f 9 have two possibility 1 , 1   &   3 , 9 Now taking m = 2   &   n = 1 we get, f 2 · 1 = f 2 · f 1 Now here f 1 = 1 , so f 2 can take all 6 numbers, Similarly, for f 5   &   f 8 there will be 6 ways each, So, total possible function will be, = 1 × 6 &#215