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You are given several pieces of paper which are as follows:

puzzle picture

(Unfortunately, a textual rendering is very difficult with this puzzle, so if someone can offer a suggestion on how to do it, that would be wonderful.)

You need to figure out what the six-character password is... Ready, set, go!

Hint 1:

Use the two outer numbers to get the middle numbers for each group of three.

Hint 2:

Read deeper into the question. Be a detective!

Hint 3:

Check the re______ h_s__r_! That's how true detectives are made!

Hint 4:

Follow this set of operations on some of the numbers you get from the triplets: ad, su, mo

Hint 5:

First step is to add the two numbers... Then, subtract some numbers that you can get after adding the two numbers...

AFTER THIS BOUNTY ENDS, I WILL POST THE ANSWER

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20
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    $\begingroup$ If we are to interpret that the central number of each triplet is derived from the two either side, then rot13(gur rknzcyrf tvira fhttrfg gung vs lbh fhogenpg gur yrsg-unaq ahzore sebz gur evtug-unaq ahzore, n artngvir erfhyg tvirf 0 naq na rira aba-artngvir erfhyg tvirf K, jvgu bgure ahzoref cebqhprq va fbzr jnl ol bqq aba-artngvir erfhygf. Fvapr gur cnffpbqr qvssreraprf ner nyy artngvir be rira aba-artngvir [0,16,-91,6,-2,12] guvf jbhyq ranoyr hf gb vtaber gur aba-0-be-K rknzcyrf naq tnva na nafjre bs KK0K0K) - I'm guessing this isn't the answer you intended...?? $\endgroup$
    – Stiv
    Nov 5, 2019 at 13:37
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    $\begingroup$ @Stiv - No, because your rule does not match what is given in the puzzle. $\endgroup$
    – Voldemort's Wrath
    Nov 5, 2019 at 18:22
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    $\begingroup$ Hint 2 suggests the solution isn't just pure mathematics, but I've spent an hour staring at the question and nothing comes to mind. $\endgroup$
    – NibblyPig
    Nov 7, 2019 at 15:45
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    $\begingroup$ @NibblyPig cryptii.com/pipes/rot13-decoder - it's just a way of preventing spoilers $\endgroup$
    – simonalexander2005
    Nov 8, 2019 at 9:31
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    $\begingroup$ @Voldemort'sWrath Hint 3 - I already found that extra clue but it didn't help $\endgroup$
    – simonalexander2005
    Nov 8, 2019 at 12:23

3 Answers 3

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I think the password is

2X3201

The method is the following

1. Add the two outer numbers to get a sum (S)
2. Seperate the sum into its last digit and the digit(s) in front of it. Get the absolute difference between these two (D).
3. Find S MOD D

As an example take $73, 5, 84$. The sum of the outer numbers is $73+84=157$. This is seperated into $15$ and $7$. Their absolute difference is $15 - 7 =8$. Now, $157$ MOD $8 = 5$, which is the middle number.

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  • $\begingroup$ I like this very much $\endgroup$
    – balazs.com
    Nov 10, 2019 at 18:43
  • $\begingroup$ I agree with balazs. I like this very much. So much so, in fact, that I will give this a shiny green check and a blasting blue upvote as well as a royal bounty! :D Congrats on the great answer! $\endgroup$
    – Voldemort's Wrath
    Nov 10, 2019 at 20:07
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Ooohh, wait! Aha! I'm going to take a shortcut,

based on just one character

and guess the password is

puzzle

Don't ask me the complete recipe, I still don't know yet!

Explanation:

The most obvious way to combine any two numbers is to try and add them. This doesn't directly render the middle number in any of the triplets. But: The fourth piece of paper, the one with the X, suggests that the middle numbers are not really numbers but characters. Indeed, 31+57=88=ascii'X'. The same holds for the third piece of paper, 9+42=51=ascii'3'. For the other papers this doesn't work. The second one yields 3+16=19, the ascii value of a non printing character, not of '3', which is 51. But the difference is 32, meaning it is shifted to the next column of the ascii table. So there seems to be some recipe. For the remaining papers the recipe appears to be more complicated, and I give up for now. Trying the simple adding recipe on the paper with the six triplets yields only one letter: 'z' on the fourth position. So we have '---z--'. The rest is just a bold guess!

Now I can't wait to press these buttons and see if the door swings open!

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    $\begingroup$ Without any explanation this doesn't answer the puzzle at all - it should probably be left as a comment instead... Also you don't need to use rot in a spoiler block - that's what the spoiler block is for :) $\endgroup$
    – Stiv
    Nov 9, 2019 at 20:22
  • $\begingroup$ How is the password a word when you are supposed to get a number from the two outside ones? $\endgroup$
    – Voldemort's Wrath
    Nov 9, 2019 at 20:22
  • $\begingroup$ @Stiv, got rid of the rot now, I was just being overly cautious. $\endgroup$
    – Zaaikort
    Nov 9, 2019 at 20:31
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    $\begingroup$ @Stiv, explanation added to my answer. $\endgroup$
    – Zaaikort
    Nov 9, 2019 at 21:22
  • $\begingroup$ The answer deserves an upvote. However, I think Hint 2 and Hint 3 render the first argument of your reasoning incorrect. $\endgroup$
    – WhatsUp
    Nov 10, 2019 at 2:36
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Answer:

21 19 51 88 157 17 49 78

Reason:

The puzzle is 6 digits long. If we add up the numbers on either side we get the following numbers. So that means that the answer must be the numbers above.

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2
  • $\begingroup$ How is this possible? Your answer is nowhere near 6 characters long. $\endgroup$
    – Voldemort's Wrath
    Nov 9, 2019 at 20:23
  • $\begingroup$ Added hint 5. Wanna try again? $\endgroup$
    – Voldemort's Wrath
    Nov 10, 2019 at 15:55

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