KöMaL (Metresis) 1894, Problem 1

Problem: 
Find all the two digit numbers $\overline{ab}$ such that $\overline{ababab}+1$ is a perfect cube.

Solution:
Let $x=\overline{ab}$ be the two digit number. Then, $101010x+1=n^3$, for some positive integer $n$. Clearly, $10^6 \leq n^3 < 10^7$, i.e. $100 \leq n \leq 215$. Moreover,
$$101010x=n^3-1.$$ Since $101010=30\cdot3367$, it's easy to see that $n^3-1 \equiv 0 \pmod{30}$ if and only if $n \equiv 1 \pmod{30}$, i.e. $n-1$ is divisible by $30$. So $n \in \{121,151,181,211\}$ and an easy check shows that only $n=211$ works. In this case, $x=93$.