Common Starting Digit For Two Powers?

Relevant wiki: General Diophantine Equations - Problem Solving

Let the common starting digit be 'm'

Then we can write, for some arbitrary integers x and y:

$m.10^x$ < $2^n$ < $(m+1).10^x$

$m.10^y$ < $5^n$ < $(m+1).10^y$

Strict inequality must hold because a power of 2 or 5 cannot be a multiple of 10.

Multiplying the two inequalities yields: $m^2.10^{x+y}$ < $10^n$ < $(m+1)^2.10^{x+y}$

Or, $m^2$ < $10^{n-x-y}$ < $(m+1)^2$

Now as m is a single digit number (something between 1 and 9) so $m^2$ and $(m+1)^2$ are both definitely something between 1 and 100. We have already seen that $10^{n-x-y}$ lies between (and not inclusive of) $m^2$ and $(m+1)^2$ . But there is only one power of 10 lying between 1 and 100 non-inclusive.

This means that $10^{n-x-y}$ = 10

So, $m^2$ < 10 < $(m+1)^2$

So 10 lies between two consecutive squares. So these squares must be 9 and 16. Thus giving m = 3.

So the answer is 3.