Let $n$ be a positive integer. Suppose that $2^n$ and $5^n$ begin with the same digit. What is the digit? 
Let $n$ be a positive integer. Suppose that $2^n$ and $5^n$ begin with
  the same digit. What is the degit?

I have a bit of difficulties to start the problem. Is anyone is able to give me a hint? Please I ask you to don't give me the solution to the question; I want to resolve by myself.
I have already know the method which is the simple mechanics; it's not interesting. The method in which we test some value of $n$, and finally we find $n=5$ respecting condition.  I would just have a more clever method. It could be any other number other than the three that meet the condition. Is anyone is able to find a rigorous and interesting method. I already know the hint $2^n 5^n = 10^n$, but I don't know what to do with this.

Assume that
$$ 2^n = (c+r_1)\cdot 10^u,\qquad 5^n = (c+r_2)\cdot 10^v \tag{1}$$
with $c\in\{1,2,3,4,5,6,7,8,9\}$ and $r_1,r_2\in [0,1)$. $(1)$ implies:
$$ 10^n = (c+r_1)(c+r_2)\cdot 10^{u+v} \tag{2}$$
hence $c$ may only be the first digit of $\sqrt{10}$, i.e. $3$, or the first digit of $\sqrt{100}$, i.e. $1$.
The second case is easy to exclude as pointed by Nate in the comments.
