A problem from a math contest about powers of two Recently I have participated in a math contest. One of the tasks really attracted me, but I still can't find the solution of it. Maybe you can help me?
Problem:

Given a natural number A which consists of 20 digits. Someone wrote the number AA...A(101 times) and removed its 11 last digits. How can I prove that this number which consists of 2009 digits isn't a power of two?


Basically try to find a power of two that yields 2009 digits. This is done with logarithms as following:
$log_{10}{2^x} = 2009$ which yields $x=6673.7535$ so $2^{6673}$ has $2009$ digits, $2^{6672}$ and $2^{6671}$ also have $2009$ digits. But we haven't finished.
Now we have to prove that the number isn't one of the aforementioned powers of $2$. The unique way is that $A$ is also a power of $2$ and has $20$ digits as well. The last power of $2$ that has $20$ digits is $2^{66}$ but surprisingly $(2^{66})^{101}$ has $2007$ digits so if we subtract $11$ digits from it it will have $1996$ digits.
