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Beginning with $2$ and $7$, the sequence $2, 7, 1, 4,7 ,4, 2 , 8,$... is constructed by multiplying successive pairs of its members and adjoining the result as next one or two members of the sequence, depending on whether the product is a one- or a two-digit number. Prove that the digit 6 appears an infinite of times in the sequence.

I kept writing the sequence a few more terms and saw that $6$ will appear a few times, I also thought about which pairs of numbers will generate one, but not as if I assume it will happen forever

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It can be verified that slightly further into the sequence you will find $\dots, 8, 2, 8, \dots$

Therefore later in the sequence you will find $\dots, 1, 6, 1, 6\dots$

Therefore later in the sequence you will find $\dots, 6, 6, 6, \dots$

Therefore later in the sequence you will find $\dots, 3, 6, 3, 6, \dots$

Therefore later in the sequence you will find $\dots, 1, 8, 1, 8, 1, 8, \dots$

Therefore later in the sequence you will find $\dots, 8, 8, 8,\dots$

Therefore later in the sequence you will find $\dots, 6, 4, 6, 4,\dots$

Therefore later in the sequence you will find $\dots, 2, 4, 2, 4, 2, 4\dots$

Therefore later in the sequence you will find $\dots, 8, 8, 8\dots$

And so on. The last 3 lines here will continue to occur in the sequence, so $6$ must appear infinitely often.

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    $\begingroup$ There may well be a much shorter proof but this is the one I found first. $\endgroup$
    – 79037662
    Oct 3 '19 at 20:30
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(Too long to be a comment)

To slightly demystify how to approach this problem, one idea is to find a never-ending sequence, which then hopefully repeats. There is nothing obvious to do, so let's start naively:

Starting with 2, 7:
2, 7, ...
1, 4, ...
4, ...
At this point, we cannot continue.

So, let's start with 2, 7, 1:
2, 7, 1, ...
1, 4, 7, ...
4, 2, 8, ...
8, 1, 6, ...
8, 6, ...
4, 8, ...
3, 2, ...
6, ...
At this point, we cannot continue.

So, let's start with 2, 7, 1, 4:
2, 7, 1, 4, ...
1, 4, 7, 4, ...
4, 2, 8, 2, 8, ...
8, 1, 6, 1, 6, 1, 6, ...
8, 6, 6, 6, ...
4, 8, 3, 6, 3, 6, ...
3, 2, 2, 4, 1, 8, 1, 8, 1, 8, ...
6, 4, 8, 4, 8, 8, 8, 8, 8, ...
2, 4, 3, 2, 3, 2, 3, 2, 6, 4, 6, 4, 6, 4, 6, 4, ...
8, 1, 2, 6, 6, 6, 6, 6, 1, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, ...
8, 2, 1, 2, 3, 6, 3, 6, 3, 6, 3, 6, 6, 2, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, ...

From here, it follows that once we get that sequence of 8, 8, 8, we get the loop as pointed out by @79037662

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