Problem Analysis #25

To find the sum of all possible values of $f(n)$ given that $g(n) = 15$, we analyze the divisors of $n$ modulo 3.

Let the prime factorization of $n$ be: $n = 3^a \prod p_i^{b_i} \prod q_j^{c_j}$ where $p_i \equiv 1 \pmod 3$ and $q_j \equiv 2 \pmod 3$.

Any divisor of $n$ that is not divisible by 3 must have the exponent of 3 equal to 0. The number of such divisors is: $N = P \cdot \prod (c_j + 1)$ where $P = \prod (b_i + 1)$ is the number of divisors formed only by primes congruent to $1 \pmod 3$.

Let $E$ be the number of divisors of $\prod q_j^{c_j}$ that are congruent to $1 \pmod 3$ (which occurs when the sum of the exponents of $q_j$ is even), and let $O$ be the number of divisors congruent to $2 \pmod 3$ (when the sum of the exponents is odd).

Thus, we have: $f(n) = P \cdot E$ $g(n) = P \cdot O$

We are given $g(n) = 15$, so: $P \cdot O = 15$ Since $P$ and $O$ are positive integers, $P$ must be a divisor of 15. Thus, $P \in \{1, 3, 5, 15\}$.

By analyzing the generating function for the exponents of $q_j$, we find that:

• If at least one $c_j$ is odd, then $E = O$.
• If all $c_j$ are even, then $E = O + 1$.

This gives two cases for $f(n)$:

1. If $E = O$: $f(n) = P \cdot O = 15$
2. If $E = O + 1$: $f(n) = P \cdot (O + 1) = P \cdot O + P = 15 + P$ Since $P \in \{1, 3, 5, 15\}$, the possible values for $f(n)$ in this case are:
   • For $P = 1$: $f(n) = 15 + 1 = 16$
   • For $P = 3$: $f(n) = 15 + 3 = 18$
   • For $P = 5$: $f(n) = 15 + 5 = 20$
   • For $P = 15$: $f(n) = 15 + 15 = 30$

All of these cases are realizable. Thus, the possible values of $f(n)$ are $15, 16, 18, 20, 30$.

The sum of all possible values of $f(n)$ is: $15 + 16 + 18 + 20 + 30 = 99$

\boxed{99}