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You are the boss at an engineering firm. Recently a rather eccentric employee named Mr. Robemer made an abrupt exit when you accused him of devoting his worktime to his hobbies. Robemer stormed out without clearing his office or, more importantly, giving the firm access to his work files. He refuses to make contact with the firm, and the employees have already spent enough time trying to crack his computer password. You have decided to take on the task; you can't afford to lose those files.

You enter his office and immediately see how eccentric he really was. You first notice two rows of four monitors he had. Robemer was a bit of a computer fanatic, but this is too much. You next find that he drew a bunch of shapes on his desk -they formed a nearly perfect octagonal-square tessellation. Doesn't look like it will just wash off. This is precisely what you were referring to when he ran off. He did not display any behavior like this when you hired him! Thoroughly confused and annoyed with the man you boot up his computer.

You check his password hint:

117 116 75 66 74 71 73

117 116 77 100 160 120 174 63 44 72 60 162 100 174 134 174 46 63 44 73

Of course he encrypted his hint. Was this a game to him, or did he have trouble remembering a complex password? He has been forgetful lately...

What is the password?

Hint 1:

Mr. Robemer seems to have favoritism towards a particular number, but it's not in the hint!

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Building off the answer by @El-Guest:

He found that the hint is ON=6<9; and ON?@pP|3$:0r@|\|&3$; by converting from octal to decimal to ASCII. His suggested password looks like computer code. If ON is a variable, 6<9 sets it to true.
The second part of the hint uses the ternary operator. If ON is true, it will return @pP|3$.
Is the password @pP|3$? (It's possible that there is a further layer of abstraction that I don't see.)

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  • $\begingroup$ Certainly, 6 < 9 == true, no? (Unless you want to turn it on its head: 6 > 9 = NO.) $\endgroup$ – M Oehm Jun 15 at 18:51
  • $\begingroup$ Yes, check your inequality. The conversion of the second line is still a little off (174 is not P). You are so close $\endgroup$ – Nordii Jun 15 at 19:12
  • $\begingroup$ Right, of course, sorry... edited. $\endgroup$ – Alex F Jun 15 at 19:32
  • $\begingroup$ You got it! It was supposed to look like apples or oranges, by the way. $\endgroup$ – Nordii Jun 15 at 19:38
  • $\begingroup$ Ohhhhhhhhhhhh... I though there might have been some obscure piece of syntax I was missing. I see it now though :) $\endgroup$ – Alex F Jun 15 at 19:58
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Since he has

2 rows of 4 computers, and an octagonal tessellation, it may be that his favourite number is 8. This is cool, because all of the numbers in the password hint appear have digits < 8...so it could be that the password hint is in octal...?

117 116 75 66 74 71 73

When converting this to decimal we get: 79 78 61 54 60 57 59. This can be converted to ASCII characters using this table to yield: O N = 6 < 9 ;

117 116 77 100 160 120 174 63 44 72 60 162 100 174 134 174 46 63 44 73

When converting this to decimal we get: 79 78 63 64 112 80 124 51 36 58 48 114 64 80 92 80 38 51 36 59. This can be converted to ASCII characters to yield: O N ? @ p P | 3 $ \$ $ : 0 r @ P \ P $\&$ 3 $ ;

Could it be possible that the password is

@pP|3$:0r@P\P&3$

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  • $\begingroup$ You are very close, but i think your conversion of the last line is flawed. I detect 6 extra chars. $\endgroup$ – Nordii Jun 15 at 18:45
  • $\begingroup$ @Nordii can you recheck? Omega Krypton’s edit seems to have inserted 6 extra characters when converting away from MathJax. Thanks! $\endgroup$ – El-Guest Jun 15 at 19:08
  • $\begingroup$ You have the right number now, but there is more meaning to what you have left. Two of your characters are also incorrectly converted. $\endgroup$ – Nordii Jun 15 at 19:31

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