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 ___________.

Solution

Maths sir · 5 steps

List the products that stay inside A first. Only 1·n and 3·3 = 9 put any condition on f.

1. Which products count. The rule applies only when m · n ∈ A. Products with 1 give 1 · n = n, and apart from those the only product that lands in A is 3 · 3 = 9. Every other product, such as 2 · 2 = 4 or 2 · 3 = 6, falls outside A and gives no condition.
2. Fix f(1). Taking m = n = 1 gives f(1) = f(1)², so f(1) = 0 or 1. Since 0 is not in A, f(1) = 1. Then f(1 · n) = f(1) · f(n) holds for every n automatically.
3. Link f(3) and f(9). Taking m = n = 3 gives f(9) = f(3)², and this square must also lie in A. Checking each value: 1² = 1 works, 3² = 9 works, while 4, 25, 64, 81 are not in A. So (f(3), f(9)) is (1, 1) or (3, 9), which is 2 choices.
4. Free values. No condition involves 2, 5 or 8, so each of f(2), f(5), f(8) can be any of the 6 elements of A. That gives 6³ = 216 ways.

   Total = 1 × 2 × 6³ = 2 × 216 = 432

5. So. The number of such functions is 432.

The common slip: 216. Keeping only f(3) = 3, f(9) = 9 and missing f(3) = 1, f(9) = 1 gives 216, not 432.

The idea to keep. For a functional equation on a finite set, first find which inputs are actually tied together. Count the choices for each linked group, let the unlinked elements go free, then multiply.

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