Does there exist a power of 2 whose last 100 digits are composed only of the digits 1 or 2? Does there exist a power of 2 whose last 100 digits are composed only of the digits 1 or 2? If so, what techniques could be used to prove its existence?


Let $q_n$ for $n>0$ be the recurring sequence of natural numbers defined by:
  \begin{align}
&q_1=1&&q_{k+1}=\frac{q_k+5^k(2-q_k\bmod 2)}2
\end{align}
  then for every $k>0$ the number $a_k=2^kq_k$ satisfy:
  
  
*
  
*$0<a_k<10^k$;
  
*the decimal digits of $a_k$ belongs to $\{1,2\}$;
  
*there exists $m\geq k$ such that $2^m\equiv a_k\pmod{10^k}$.
  

While (1) and (2) are easily proved by induction, to prove (3), note that $5\nmid q_k$, for otherwise $10\mid a_k$ but this would implies that the last digit of $a_k$ is zero - a contradiction.
Since $2$ is a primitive root modulo $5^k$, there exists $n>0$ such that
$$2^n\equiv q_k\pmod{5^k}$$
Then multiplying by $2^k$, we get
$$2^{n+k}\equiv a_k\pmod{10^k}$$

To generate the recurrence above: assume $2^m\equiv a_k\pmod{10^k}$ with $m\geq k$ and $a_k=2^kq_k$ and let $2^n\equiv 10^kd+a_k\pmod{10^{k+1}}$ with $d\in\{1,2\}$ and $n\geq k+1$.
Dividing by $2^k$ we get $2^{n-k}\equiv 5^kd+q_k\pmod{2\cdot 5^{k+1}}$, hence reducing modulo $2$, we get $d\equiv -q_k\pmod 2$.
Since $d\in\{1,2\}$, we have $d=2-q_k\bmod 2$.
Moreover,
$$2^n\equiv 2^k(5^kd+q_k)=2^{k+1}q_{k+1}\pmod{10^{k+1}}$$
where $q_{k+1}=\frac{q_k+5^kd}2$.
