Let f be a function such that f (mn) = f (m) f (n) for every positive integers…

A function f defined on positive integers with f(mn) = f(m) × f(n) for every pair of positive integers m, n (not only coprime ones) is called completely multiplicative. Two facts follow directly from this: f(1) must equal 1, and if a number is written as a product of prime powers, f of that number equals the product of f(prime) raised to the matching power for each prime.

Applied at the primes 2 and 3, this is exactly what is needed here: find f(2) and f(3) from f(24) = 54, then reuse them for f(18).

1. Find f(1): since f(1×1) = f(1)×f(1), we get f(1) = f(1)2, i.e. f(1)(f(1) − 1) = 0. As f(1) is a positive integer, f(1) = 1.

2. Write 24 in prime powers: 24 = 23 × 3. So f(24) = f(2)3 × f(3) = 54.

3. Factor 54 the same way: 54 = 2 × 27 = 2 × 33.

4. Since f(1) < f(2), f(2) must be an integer greater than 1. Try f(2) = 3: f(2)3 = 27, so f(3) = 54 ÷ 27 = 2, a positive integer — this is consistent.

5. Check there is no other option: f(2) = 2 gives f(2)3 = 8, and 54 ÷ 8 is not an integer; f(2) = 6 gives f(2)3 = 216, already larger than 54. So f(2) = 3, f(3) = 2 is the only valid pair.

6. Write 18 in prime powers: 18 = 2 × 32. So f(18) = f(2) × f(3)2 = 3 × 22 = 3 × 4 = 12.

Cross-check: f(24) = f(2)3 × f(3) = 33 × 2 = 27 × 2 = 54, matching the given value, and f(1) = 1 < f(2) = 3 as required — both conditions hold.

So f(18) = 12.