For every positive integer $n$ let $\tau (n)$ denote the number of its positive factors. Determine all $n \in N$ that satisfy the equality $\tau (n) = \frac{n}{3}$
Problem
Source: 2011 Romania JBMO TST 4.1
Tags: factors, positive, number theory
Snowyowl2005
18.06.2020 01:58
Note that $\tau(n) \leq 2 \sqrt n$, and so: \[\frac{n}{3} \leq 2 \sqrt{n}\]\[n \leq 18\]We can check manually that the only $n$ that work are $\boxed{n = 9}$ and $\boxed{n = 18}$
mvishal
20.05.2021 07:21
Note that \[\Big(\frac{1}{2}\Big)\Big(\frac{2}{3}\Big)=\Big(\frac{4}{8}\Big)\Big(\frac{2}{3}\Big)=\Big(\frac{3+1}{2^3}\Big)\Big(\frac{1+1}{3^1}\Big)=\frac{(3+1)(1+1)}{2^3\cdot3^1}\Rightarrow n=2^3\cdot3^1=\boxed{24}.\]Hence, $n=24$ works too.
Here, I present a reliable and easy solution.
First, we rearrange the above equation to make use of the multiplicative property.
\[\frac{\tau(n)}{n}=\frac{(e_1+1)(e_2+1)\cdots(e_k+1)}{p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}}=\prod_{i=1}^k \frac{(e_i+1)}{p_i^{e_i}}=\frac{1}{3}.\]Note that exponential functions grow faster than linear functions. Let's analyse small values of $\frac{e+1}{p^e}$.
\begin{tabular}{|c|c|c|c|c|}
\hline
& $p=2$ & $p=3$ & $p=5$ & $p=7$ \\
\hline
$k=1$ & $\color{green}{\frac{1+1}{2^1}=1}$ & $\color{green}{\frac{1+1}{3^1}=\frac{2}{3}}$ & $\color{green}{\frac{1+1}{5^1}=\frac{2}{5}}$ & $\color{red}{\frac{1+1}{7^1}=\frac{2}{7}}$ \\
\hline
$k=2$ & $\color{green}{\frac{2+1}{2^2}=\frac{3}{4}}$ & $\color{green}{\frac{2+1}{3^2}=\frac{1}{3}}$ & $\color{red}{\frac{2+1}{5^2}=\frac{3}{25}}$ & $\color{red}{\vdots}$ \\
\hline
$k=3$ & $\color{green}{\frac{3+1}{2^3}=\frac{1}{2}}$ & $\color{red}{\frac{3+1}{3^3}=\frac{4}{27}}$ & $\color{red}{\vdots}$ & \\
\hline
$k=4$ & $\color{red}{\frac{4+1}{2^4}=\frac{5}{16}}$ & $\color{red}{\vdots}$ & & \\
\hline
\end{tabular}All the red boxes have values of $\frac{e+1}{p^e}$ greater than $\frac{1}{3}$, so we can disregard these boxes. This reduces our search to just the green boxes above. Inspection gives us that the only solutions are:
\[\frac{1}{3}=\frac{3}{9}=\frac{(2+1)}{3^2}\Rightarrow n=3^2=\boxed{9}.\]\[(1)\Big(\frac{1}{3}\Big)=\Big(\frac{2}{2}\Big)\Big(\frac{3}{9}\Big)=\Big(\frac{1+1}{2^1}\Big)\Big(\frac{2+1}{3^2}\Big)=\frac{(1+1)(2+1)}{2^1\cdot3^2}\Rightarrow n=2^1\cdot3^2=\boxed{18}.\]\[\Big(\frac{1}{2}\Big)\Big(\frac{2}{3}\Big)=\Big(\frac{4}{8}\Big)\Big(\frac{2}{3}\Big)=\Big(\frac{3+1}{2^3}\Big)\Big(\frac{1+1}{3^1}\Big)=\frac{(3+1)(1+1)}{2^3\cdot3^1}\Rightarrow n=2^3\cdot3^1=\boxed{24}.\]
jasperE3
20.05.2021 08:06
Since $\tau(n)\le2\sqrt n$, we have $n\le6\sqrt n$, so $n\le36$. Noting that primes and semiprimes other than $9$ cannot work, we manually check, yielding only $n\in\boxed{\{9,18,24\}}$.