This conclusion can be proved in many different ways. Here is a solution using modular equations. Let $x=10^k$. Then $$\left\{ \begin{array}{ccc} x = 10^k\equiv 0 \pmod{2}&\implies & x\equiv -2\pmod{2}\\ x = 10^k \equiv 1\pmod{3}&\implies &x\equiv -2\pmod{3} \end{array} \right.$$

This means the solution is $2$ less than the least common multiple of $2$ and $3$ which is $6$. Or $$10^k\equiv -2\equiv 4\pmod{6}$$