Numeric sequence [duplicate]

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

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.

• There may well be a much shorter proof but this is the one I found first. Oct 3 '19 at 20:30

(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.

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.