Some years ago I submitted a problem to Crux Mathematicorum, but sadly I still can't see the possible solutions because I'm not a subscriber. So, I decided to post here my problem and see if there is a solution different than mine.
Here is the problem: if $\sigma(.)$ and $\tau(.)$ denote the sum and the number of divisors of a positive integer, respectively, then find all $n\in \mathbb{Z}^{+}$ such that $\tau(\sigma(n))=n$.
Here's an alternative, using $\tau(n)\le\sqrt{3n}$ and
$$\sigma(n)\le n\left(1+{1\over2}+\cdots+{1\over n}\right)\lt n(1+\ln n)$$
If $n=\tau(\sigma(n))$, then
$$n\le\sqrt{3\sigma(n)}\lt\sqrt{3n(1+\ln n)}$$
or
$$n\lt3(1+\ln n)$$
It's easy to check that this final inequality is satisfied only for $n\le9$, which leaves a small amount of computation to find $n=1$, $2$, $3$, and $6$ as the only solutions.