Assume FTSOC that the decimal representation of $a$ is eventually periodic (terminating is periodic with period $1$). Now let $2^n = a_1\cdot 10^{m-1}+a_2\cdot 10^{m-2}+\dots$ be the decimal representation of $2^n$. Then, divide by $10^{m-k}$, and take the floor function we see that $\left\lfloor\frac{2^n}{10^{m-k}}\right\rfloor = a_1\cdot10^{k-1}+\dots+a_k$. But the expression inside the floor turns out to be $\frac{10^{k-1}\cdot 2^n}{10^{m-1}} = \frac{10^{k-1}\cdot 2^n}{10^{\lfloor n\log{2}\rfloor}} = 10^{k-1}\cdot 10^{\{n\log{2}\}}$, but since $\log{2}$ is irrational, if the periodicity of the decimal representation is $T$, then for any given $a$ we can find $b$ such that $\{n\log{2}\}$ lies in an interval that is a subset of $[0, 1)$ of our choice so that $\lfloor 10^{k-1}\cdot 10^{\{n\log{2}\}}\rfloor$ is an integer of our choice in $(10^{k-1}, 10^k)$.