Let S = 1, 2, 3, 4, 5, 6, 7. Then the number of possible functions f : S → S
such that f(m . n) = f(m) . f(n) for every m, n ∈ S and m . n ∈ S is equal to _____________.

Solution

Maths sir · 4 steps

Only a few products stay inside S, so fix f(1), then split on f(2) and count.

1. Products inside S. The condition applies only when m · n ≤ 7. Apart from 1 · n = n, the only such products are 2 · 2 = 4 and 2 · 3 = 6. So 5 and 7 never appear in a constraint, and f(5), f(7) can each be any of the 7 values.
2. Fix f(1). Taking m = n = 1 gives f(1) = f(1)², so f(1) = 1 because 0 ∉ S. With f(1) = 1, every condition f(1 · n) = f(1) f(n) holds automatically.
3. Split on f(2). f(4) = f(2)² must lie in S, so f(2) = 1 or 2. Also f(6) = f(2) f(3) must lie in S, which limits f(3). Here f(4) and f(6) are then fixed.

   f(2)	f(4)	Allowed f(3)	f(5), f(7)	Functions
   1	1	any of 7, and f(6) = f(3)	7 × 7	7 × 49 = 343
   2	4	1, 2, 3 only, since 2f(3) ≤ 7	7 × 7	3 × 49 = 147

4. So. The total number of functions is 343 + 147 = 490.

The common slip: 686. Letting f(3) be anything when f(2) = 2 gives 343 + 343 = 686, but then f(6) = 2f(3) can exceed 7, so only f(3) = 1, 2, 3 work.

The idea to keep. In a functional equation on a finite set, list exactly which products stay inside the set. Values that appear in no product are free, and the rest are fixed or restricted by the few that do.

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