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It is true that every power of $2$ of the form $2^{6+10x}$, $x\in\mathbb{N}$, has $6$ as one of its digits. Something more is true, the last two digits are either $64$ or $36$. The OP suggests that "...Beyond that the numbers are so large that they seem to all contain a 6 as part of their digits." This leads to the following related questions.

Q.1. Is there a natural number $k$ such that $2^n$ has $6$ as one of its digits for all natural numbers $n\ge k$ ?

Q.2. Is there a natural number $k$ and a natural number $j$ such that the last $j$ many digits of $2^n$, for $n\ge k$, would cycle through only finitely many possibilities, each of which has a $6$ as one of its digits? I believe that formally this is asking the following. Are there triples $(k,j,\mathcal F)$ where $k,j$ are natural numbers and $\mathcal F$ is a finite set of numbers each with $j$ many digits one of which is necessarily $6$ (leading zeroes allowed) such that $2^n\mod 10^j \in\mathcal F$ whenever $n\ge k$ ? ("Cycling" also suggests that there is a particular order under which the elements of $\mathcal F$ appear, as $n$ increases, and this order repeats, that is say $\mathcal F=\{F_0,F_1,...,F_{m-1}\}$ and $2^n\mod 10^j =F_{(n \mod m)}$ whenever $n\ge k$, where each $F_t$ has $6$ as one of its digits. So there are two versions of Q.2, with or without cycling. Here $2^n\mod 10^j$ is of course understood as a function, $\mod(2^n,10^j)$, with values in the set $\{0,1,...,10^j-1\}$.)

Q.3. Which is, on its own, several more questions, namely the variations of Q.1 and Q.2 when we use a different digit instead of $6$? (And even more variations, what if we consider a different base than $2$?)

To keep it focused, Q.1 is the "official" question posted here, but clearly Q.2 is very closely related (and perhaps even more interesting), and Q.3 seems related too (but feels too broad for now, perhaps should be posted separately). (Of course having this question depend on decimal representation makes is a bit lacking in motivation, but it might be interesting anyway.)

Edit. Collecting some of the comments posted during the first two hours, plus some numerical data.

I agree with @LaBird that the answer to Q.1 is likely yes, with $k=94$. There is no 6 in $2^{93}=9903520314283042199192993792\ \ \ $ but my computer confirmed that $2^n$ has a 6 for $n$ from $94$ up to at least $216927\ $. LaBird also provided a link to a related question a special case of which suggests that $2^n$ has all digits for $n\ge169$ (confirmed for $n$ up to $10000$)
Biggest powers NOT containing all digits.

My computer also tells me that the last $60$ digits of $2^{160021}$ have no $6$, but the last $100$ do.
The last $100$ digits of $2^{52613}$ have no $6$ (though $2^{52613}$ itself does).
The last $101$ digits of $2^{183137}$ have no $6$ (though $2^{183137}$ itself does).

@Soke says we know that the $m$-many last digits do cycle with period $4 \cdot 5^{m-1}$ (I didn't know and have yet to verify it for myself or find a reference, but I have no reason to doubt it). Does that make Q.1 appear easy (may boil down to a tedious verification)? On the other hand @Elaqqad suggests similarity to a hard problem of Erdős open since 1979, that the base-$3$ representation of $2^n$ contains a $2$ for all $n>8$. I found the following two related links:
http://mathworld.wolfram.com/Ternary.html and
https://terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-littlewood-offord-theory-and-powers-of-2-and-3/

@Elaqqad also provided a link to MSE question discussing the related conjecture that all powers of $2$ greater than $2^{86}$ have a zero in their base $10$ representation
Status of a conjecture about powers of 2

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  • $\begingroup$ (My point of view of the question) This is a good problem, but I think the answer will be very long which is against MSE politics. I think that the questions $Q.3$ and $Q.2$ are "open-ended, hypothetical questions". (the first question is a very good question if asked seperately $\endgroup$ – Elaqqad Mar 21 '15 at 15:19
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    $\begingroup$ We know that the $m$-th to last digit cycles with period $4 \cdot 5^{m-1}$, so, if proposition 1 is true, it may just boil down to a very tedious hand-check of these digit cycles. $\endgroup$ – MCT Mar 21 '15 at 16:32
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    $\begingroup$ I wrote a program to calculate $2^n$ for integer $n$ up to $30000$ and check for the presence of each of the $10$ digits, and seems the answer to Q1 is $94$. Also, another observation is that the smallest $k$ such that $2^n$ contains all $10$ digits for all $n \ge k$ is $k=169$. This matches the description and comments in this question: math.stackexchange.com/questions/627409/… $\endgroup$ – LaBird Mar 21 '15 at 16:42
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    $\begingroup$ This seems to be similar to (I'd venture to say as hard as) a problem of Erdős open since $1979$, that the base-3 representation of $2^n$ contains a $2$ for all $n>8$. $\endgroup$ – Elaqqad Mar 21 '15 at 16:51
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    $\begingroup$ See also another formulation here for the digit $0$, Status of a conjecture about powers of 2 $\endgroup$ – Elaqqad Mar 21 '15 at 16:57
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Q1: This is easy to state but to me seems hard to proof:

Each $i$-th place $d_i^{(n)}$ of $2^n=(d_{m-1}^{(n)}\cdots d_0^{(n)})_{10}$ develops with increasing $n$ similar to a pseudo random number generator, generating digits from $\{0,\ldots,9\}$. The length $m$ of this representation increases, introducing more generating places, so the probability that one of the $m$ places generates a $6$ is growing. Or the other way round: for large $n$ and thus large $m$ it becomes more and more unlikely that none of the $m$ places generates a digit $6$. All these places are coupled linear congruential generators $$ d_i^{(n+1)} = (2 d_i^{(n)}+c_i^{(n)})\mbox{ mod }10 $$ with different initial values and different couplings via the carry $c$ between places. Each can have a different initial run and a different cycle of generated digits with different cycle length. I am not sure if one can analyze them combinatorically to come up with a lower limit $k$ For $n$, or if this line of thought stays probabilistic.

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  • $\begingroup$ I have yet to read your answer about Q.1, but concerning the part you have about Q.2, I do not get it. Certainly, there are finitely many remainders mod $10^j$, for a fixed $j$, but do they all have a $6$? Well, I also read the part about Q.1, yes, probabilistically one might expect positive answers. $\endgroup$ – Mirko Mar 21 '15 at 17:07
  • $\begingroup$ I might understand your Q2 wrong, so I remove that bit. $\endgroup$ – mvw Mar 21 '15 at 20:59

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