An algebra problem by Rahil Sehgal

We can see that $f(1 \cdot 1) = f(1) \cdot f(1) = f(1) ^2$ , implying $f(1) = 1$ . Thus, the smallest values $f(x)$ can take are $2$ and $3$ .

Now, $f(999) = f(3 \cdot 3 \cdot 3 \cdot 37) = f(3) ^3 \cdot f(37)$ . Now, since we want to minimize this product, we minimize $f(3)$ and $f(37)$ . We can see that either $f(3) = 2, f(37) = 3$ or $f(3) = 3, f(37) = 2$ , but trying those out gives us the smaller value when $f(3) = 2, f(37) = 3$ , giving us $f(999) = f(3) ^3 \cdot f(37) = 2^3 \cdot 3 = 24$ as a possible minimum.

Now, we just have to prove that such a function exists. Note that if we set

$f(p) = p$ if $p$ is prime and not equal to $2$ , $3$ , or $37$

$f(3) = 2, f(2) = 37, f(37) = 3, f(mn) = f(m)f(n)$ ,

We can solve for the value of $f(x)$ uniquely using all of the givens. This implies that there exists a function for which $f(999) = 24$ , and thus, the minimum of $f(999)$ is $\boxed{24}$ .

Since $f(1) = f(1)^2$ , we must have $f(1) =1$ . Thus $f(3),f(37) \ge 2$ , and $f(3) \neq f(37)$ . Since $999 = 3^3 \times 37$ , we deduce that $f(999) \ge 2^3 \times 3 = 24$ .

Let $p_1,p_2,p_3,\ldots$ be a listing of all the prime numbers, and let $q_1,q_2,q_3,\ldots$ be a sequence of distinct primes. Then we can define a function with the desired property such that $f(p_j) = q_j$ for all $j \ge 1$ . For this function $f(999) = f(3)^3f(37) = f(p_2)^3f(p_{12}) = q_2^3q_{12}$ . We can achieve $f(999) = 24$ by choosing $q_2=2$ and $q_{12}=3$ (for example, we could have $q_1=37$ , $q_2=2$ , $q_{12}=3$ and $q_n=p_n$ for all other $n$ ). Thus the minimum value of $f(999)$ is $\boxed{24}$ .