Problem 1555

If $n \in \mathbb{N}$ satisfies the condition $\tau(n) = \frac{n}{3}$, then $3 \mid n$. Put $n = 3k$, $k \in \mathbb{N}$. If $k$ is even then $\frac{k}{2} = \frac{n}{6}$ is a factor of $n$. Even if all the positive numbers smaller than $\frac{n}{6}$ are factors of $n$ and the numbers $\frac{n}{5}, \frac{n}{4}, \dots, \frac{n}{1}$ are also factors of $n$, we have $\frac{n}{3} = \tau(n) \le \frac{n}{6} + 5$. Hence $n \le 30$. Checking the numbers $6, 12, 18, 24$ and $30$ we find that $18$ and $24$ satisfy the desired condition.

In the case when $k$ is odd we can proceed similarly, obtaining $n = 9$, or, alternatively, we can use the fact that a number having an odd number of positive factors is a perfect square. If $n = m^2$, then $n$ has at most $m + 1$ factors, hence $m + 1 \ge \frac{m^2}{3}$. We obtain that $m \le 3$ and the conclusion.

In conclusion, the problem admits three solutions: $9$, $18$ and $24$.

Since $3 \mid n$, it follows that $n$ has a prime factorization of the form $n = 3^a \cdot p_1^{\alpha_1} \cdot \dots \cdot p_j^{\alpha_j}$ and the number of its positive factors is $\tau(n) = (a+1)(\alpha_1+1)\dots(\alpha_j+1)$. The condition $\tau(n) = \frac{n}{3}$ leads to $3^{a-1} \cdot p_1^{\alpha_1} \cdot \dots \cdot p_j^{\alpha_j} = (a+1)(\alpha_1+1)\dots(\alpha_j+1)$. Since $p_i^{\alpha_i} \ge 2^{\alpha_i} \ge \alpha_i + 1$, in order for $n$ to satisfy the equation from the statement, it is necessary that $a+1 \ge 3^{a-1}$, hence $a=1$ or $a=2$.
If in the prime factorization of $n$ there is a prime $p_i > 3$, then $p_i^{\alpha_i} > 4^{\alpha_i} \ge 2\alpha_i + 2$ and the equality $\tau(n) = \frac{n}{3}$ can not take place. If $a=1$ then $n = 3 \cdot 2^m$ and the equality $\tau(n) = \frac{n}{3}$ reduces to $2(m+1) = 2^m$. We find $m=3$ (for $m \ge 4$ we have $2^m > 2m+2$), hence $n=24$.
Similarly, if $a=2$ then $n = 3^2 \cdot 2^m$ and from $3(m+1) = 3 \cdot 2^m$ we obtain $m \in \{0,1\}$, i.e. $n \in \{9,18\}$.