19
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This is what the diagram on the machine looked like:

sequencer

And this is the output that was in the box.

 '297437431016338',
 '0330628069082',
 '26248092373710',
 '7600101586999295',
 '673949897',
 '519961796487351',
 '99157454734',
 '44752501089021',
 '06735883184',
 '7351473960715',
 '861811447525',
 '74215796196458',
 '200708524570',
 '184574951378255',
 '5124776841611395',
 '161139586018',
 '237371095093751',
 '942213295895',
 '9084262013679422',
 '20735427280960',
 '44828805417',
 '78434900766902',
 '6397600482033',
 '28425934200448',
 '6793620054046',
 '00401901710',
 '245370289410171',
 '992952842593',
 '0943487058936',
 '05417002833',
 '379253772329743',
 '9693140354640',
 '6496719513382',
 '3387546081209801',
 '04003273563985',
 '07669025586181',
 '37517427279894',
 '89523366810',
 '45055751791437',
 '5245706680969',
 '6071519064967',
 '0022124205655287',
 '734120911',
 '505458358546916',
 '9021437664963792'

I wonder what the previous operator did?

Hint:

The machine is meant to be analogous to a different kind of sequencer

Hint 2:

The input that the operator gives is relatively short. The diagram indicates the input format.

Hint 3:

@Epiksalad has found the two possible candidates for the interim step (long line at the top). Try factorizing them and see if that gives you any ideas...

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  • 5
    $\begingroup$ Why is this ringing so many bells, but I'm not able to put a finger on it. $\endgroup$ – Don Thousand Apr 5 at 22:05
  • 2
    $\begingroup$ This is so confusing... $\endgroup$ – Xilpex Apr 5 at 22:17
  • $\begingroup$ This is basically an analogy for how a DNA sequencer works. Apart from the initial operation. $\endgroup$ – Dr Xorile Apr 17 at 16:35
7
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The strings of numbers combine to form a long string of numbers.
There are many pairs of strings which end in the same 3- or 4-digit sequence like so: $a_{1}...a_{2}a_{3}A_{1}A_{2}A_{3}$ and $b_{1}...b_{2}b_{3}A_{1}A_{2}A_{3}$.
One of these strings may be listed flipped:
$a_{1}...a_{2}a_{3}A_{1}A_{2}A_{3} \rightarrow A_{3}A_{2}A_{1}a_{3}a_{2}...a_{1}$
The two strings can be concatenated (flipped if necessary) and the shared digits $A_{1}A_{2}A_{3}$ removed. Repeating this process many times yields one long string.


In my first attempt at an answer I mistakenly forgot to copy the last 8 strings listed in the question. This was my result from concatenation:
'6496719513382007145088284400243952482592999685101006735883184574951378255',
'3792537723297437431016338754608120980105257441181685520966700943487058936537230040190171014982073542728096082603302840067936200540464530413969',
'59859231224976310262480923737109509375174272798949376',
'4374547519961796487351473960715',
'810685931161486774215796196458',
'200708524570'

Old answer: "What these mean I can't figure out. Tried decrypting them using various means. Theres a 31bit almost? key in there for a reason"

New:
Adding the missing 8 strings to this mix allows me to combine the above result into one long string: 119021437664963792537723297437431016338754608120980105257441181685520966700943487058936537230040190171014982073542728096082603302840067936200540464530413969086607542580700221242056552873159475481388537600101586999295284259342004482880541700283315917694609151706937415378469716991574547341971575505458358546916975124776841611395860186633259859231224976310262480923737109509375174272798949376.
This number is factorizable as $2^{500}$ * $3^{500} $= $6^{500}$.

The function of the sequencer is apparently to take the input $a \square b$, convert it to $a^{b}$, then repeatedly cut the result into segments.
Did the operator input 6 and 500?

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  • 2
    $\begingroup$ How did you combine the sequences? $\endgroup$ – Dr Xorile Apr 10 at 16:15
  • 1
    $\begingroup$ The ends of the sequences are the same, but many segments are flipped. Converting to Hexadecimal one sequence ends in 13 0s which seems too many too be a coincidence $\endgroup$ – Artemmm Apr 10 at 21:09
  • 1
    $\begingroup$ The hexadecimal thing is not a coincidence but rather an unexpected by-product. $\endgroup$ – Dr Xorile Apr 15 at 16:28
  • 1
    $\begingroup$ So those zeros were 55 of 500 powers of 2. Funny how that turned out $\endgroup$ – Artemmm Apr 16 at 21:46
3
+50
$\begingroup$

Partial answer

As Artem Lugin did, I combined the sequences to form a larger sequence. If you look carefully, there are several pairs of strings such that the end of one string is the same as the beginning of another. Further, you have to reverse some strings for them to join up with another.

For example,

'379253772329743' and '297437431016338'. Another example is '74215796196458' and '505458358546916' where the second string has to be reversed. On combining all the parts like these and deleting the common parts between each pair, it leads to one very long string: 673949897272471573905901737329084262013679422132958952336681068593116148677421579619645853854505575179143745475199617964873514739607151906496719513382007145088284400243952482592999685101006735883184574951378255650242122007085245706680969314035464045002639760048203306280690827245370289410171091040032735639850784349007669025586181144752501089021806457833610134734792327735297369466734120911

However,

It could also be that the required string is the reverse of the above string: 119021437664963792537723297437431016338754608120980105257441181685520966700943487058936537230040190171014982073542728096082603302840067936200540464530413969086607542580700221242056552873159475481388537600101586999295284259342004482880541700283315917694609151706937415378469716991574547341971575505458358546916975124776841611395860186633259859231224976310262480923737109509375174272798949376

I'm not able to figure out what to do next. Tried decrypting it via various means but nothing seems to be working. Any hints?

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  • $\begingroup$ Great job! I'll add another hint in the question. $\endgroup$ – Dr Xorile Apr 13 at 20:24
  • $\begingroup$ Now I see I didn't do a good copy/paste. My answer is missing the last 8 strings $\endgroup$ – Artemmm Apr 16 at 21:08
  • $\begingroup$ I've awarded the bounty. But you're on the right track. Keep going. I've added more clues. $\endgroup$ – Dr Xorile Apr 16 at 21:40
  • $\begingroup$ Ah, after reading the correct answer, I wasn't even close. I doubt I would have got it any time soon. Good problem though :D $\endgroup$ – Epiksalad Apr 17 at 17:38

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