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?

 A: First of all, note that instead of removing the last $11$ digits, we might as well remove the first $11$ digits — this change just corresponds to permuting the digits of $A$. For example, if $A=11223344556677889900$, the entire number will look like
$$
\underbrace{11223344556677889900}_{\text{repeated 100 times}} \,112233445
$$
which could just as well be written as
$$
112233445\,\underbrace{56677889900112233445}_{\text{repeated 100 times}}
$$
(You could worry that this isn't quite true if $A$ has a $0$ in the wrong place in its decimal expansion, but in fact our proof will work regardless of whether $A$ is "really" a $20$-digit number or a shorter nonzero number padded with zeroes on the left.)
This means that our number is:
$$10^{2000} * \text{a $9$-digit number} + \underbrace{AA\dots A}_{100\text{ times}}$$
The first term in this sum is a multiple of $2^{2000}$. So if the whole sum is to be a power of $2$ the second term must also be a multiple of $2^{2000}$ (since the entire sum would be $2^{\text{something a lot bigger than 2000}}$).
But
$$
\underbrace{AA\dots A}_{100\text{ times}}=A * (1\underbrace{\underbrace{00\dots0}_{19\text{ times}}1\underbrace{00\dots0}_{19\text{ times}}\dots\underbrace{00\dots0}_{19\text{ times}}1}_{\text{99 times}})
$$
which is $A$ times an odd number. Note that $A$ is at most a $20$-digit number, and $2^{2000}$ has a lot more than $20$ digits. So $A$ cannot be a multiple of $2^{2000}$, which means $A$ times an odd number also cannot be a multiple of $2^{2000}$ — meaning the entire sum is not a multiple of $2^{2000}$ and hence not a power of $2$.
A: 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.
