==> arithmetic/digits/power.two.s <==
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.