Let $\tau(n)$ denote the number positive divisors of $n$. Given that $\tau(10x) = x$, find $x$ where $x$ is an integer. The problem:
Let $\tau(n)$ denote the number positive divisors of $n$.
Given that $\tau(10x) = x$, find $x$ where $x$ is an integer.

I'm using this specific example to find a technique to use in related results. I can find the value (x = 18) by trial-and-error, but I need a more rigorous way of doing this. I feel like I'm over-looking something simple. Any tips are appreciated.
 A: Let $r=\tau(n)$. If $1=d_1<d_2<d_3<\cdots < d_r = n$ the positive divisors of $n$.
Consider the table
\begin{array}{}
d_1\cdot d_r&=n\\
d_2\cdot d_{r-1}&=n\\
d_3\cdot d_{r-2}&=n\\
\hfil{\vdots}\\
d_{\lfloor r/2\rfloor}\cdot d_{\lceil r/2\rceil}&=n
\end{array}
Since each factor on the left side $d_i$ with $i = 1,\ldots ,\lfloor r/2\rfloor$ is less than or equal to $\sqrt{n}$ the table has at most $\sqrt{n}$ rows. It follows that $$\tau(n) < 2\sqrt{n}$$
(I leave the details to you).
Now we use this inequality to this equation $\tau(10x) = x$.
We have $$x = \tau(10 x) < 2\sqrt{10x}$$
It follows that $x < 40$, and now you just have to check a small number of values for $x$.
Edit: With more effort one can prove that $\tau(n) \le \sqrt{3n}$, this reduce the search to $x \le 30$.
A: Let $x=p_1^{a_1}...p_n^{a_n}$ where $p_i$'s are primes and $a_i$'s positive integers. Number of divisors of $x$ is given by
$$(a_1+1)(a_2+1)...(a_n+1)$$
So, there are 4 scenarios for divisors of $10x$. (Like $x$ containing only $2$ or only $5$ or non of it or both in its prime divisors).
Lets examine the case where 2 and 5 don't divide $x$. Then we should have
$$x=p_1^{a_1}...p_n^{a_n}=(a_1+1)(a_2+1)...(a_n+1)(2)(2)=\tau(10x)$$
**Not an answer but this is whole I got. Maybe this can be helpful.
A: Let r(x) = tau(10x) / x. You are looking for r(x) = 1, we are also interested in when r(x) >= 1.
We have r(1) = 4. Multiplying an odd x with powers of 2 multiplies r(x) by 3/4, 4/8, 5/16, 6/32 etc. Multiplying by a power of 5 multiplies r(x) by 3/10, 4/50 etc. Multiplying by another prime p multiplies r(x) by 2/p, 3/p^2, 4/p^3 etc. Since we start with r(1)=4, and want r(x) >= 1, the only possible factors are 2, 4, 8, 5, 3, 9, and 7, everything else will give r(x) < 1.
So values r(x) >= 1 are: r(1)=4, r(2)=3, r(4)=2, r(8)=1.25, r(5)=1.2, r(3)=8/3, r(6)=2, r(12)=4/3, r(9)=4/3, r(18)=1, and r(7)=8/7, and clearly no larger ones.
A: First, note that for $N$ large, $\tau(N) << N$, and even $\tau(10N) < N$. So there are not so many values to check for $N$ at all. Indeed, check for yourself first that, in your equation above, that
$p_i^{a_i} > (a_i+1)$, and in fact, $p^{a_i}_i > 2(a_i+1)$ for say $p_i \ge 7$ or $a_i > 3$, and $p^{a_i} > 10(a_i+1)$ for $a_i \ge 5$. So there
cannot be more than a few distinct $p_i$s in the factorization of $x$, and none of the $a_i$s can be larger than $4$. For otherwise $\tau(10x)$ will be less than $x$. Thus, for the inequality $x = \tau(x)$ and even $x = \tau(10x)$ to hold, the integer $x$ cannot be too large.
