Let v = log to base 10 of 2. Then v is irrational.
Let w = log to base 10 of these 9 digits.
Since v is irrational, given epsilon > 0, there exists some natural number n such that
{w} < {nv} < {w} + epsilon
({x} is the fractional part of x.) Let us pick n for when
epsilon = log 1.00000000000000000000001.
Then 2^n does the job.