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I've been an amateur puzzle maker for a long time. I've never really found an audience and someone suggested that I check this place out. So my puzzles are kinda ARG style. Let's see if you can figure out this one and uh please leave feedback below as well heheh. All I can say is that the answer will be clearly stated.

I'm gonna spice up the deal with the first person getting this getting a steam game out of a couple choices!

The puzzle!

the puzzle

As you might be able to see, there are 3 components to this, separate them.

The spiralling numbers are for aesthetic but the crossed out boxes are not The numbers in the 4 by 4 grid follow a pattern and the number on the bottom right wouldn't be 16 The pattern for the 4 by 4 square is indeed 1 being one space away from 2, 2 being 2 spaces away from 3. Etc. Wouldn't it be fun to generalize the problem? The playfair square goes left to right, top to bottom

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  • $\begingroup$ I don't think the puzzle-creation tag is applicable here, but that said, this puzzle looks nice! $\endgroup$ – Wen1now Sep 28 '17 at 2:22
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    $\begingroup$ What does "ARG style" mean? $\endgroup$ – Jamal Senjaya Sep 28 '17 at 2:29
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    $\begingroup$ @JamalSenjaya Alternate Reality Game, maybe? $\endgroup$ – CinCout Sep 28 '17 at 6:48
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    $\begingroup$ Indeed that's what I mean. Also I felt puzzle creation was applicable due to me wanting advice. $\endgroup$ – Quadraxus Sep 28 '17 at 11:51
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    $\begingroup$ No not quite, would you guys like a hint? $\endgroup$ – Quadraxus Sep 29 '17 at 23:20
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Update, but still not even a partial answer, I'm afraid. I'm just thinking aloud and throw out some ideas.

What is the missing number?

AstroMax has found the actual pattern: the number n comes n places after the previous number, n − 1. This pattern uases the rows as lines and it wraps when the bottom right is reached.

This pattern can be continued until a number is placed in the last cell, which is 31. The next step would place a 32 in the same box, and by now, every cell has exactly two numbers:

two numbers per cell

The pattern can be continued infinitely, of course.

What is the meaning of the diagonal bars?

There are sixteen cells overall and ten of them are divided by a diagonal bar. That makes six full cells and 20 half-cells. That's 26 cells in total, which conveniently is the number of English letters. Hmm.

My guess was that each cell referred to a letter. Meanwhile, the OP has stated that the grid is a Playfair square. Typically, Playfair uses a 5×5 grid for the 26 english letters. The excess letter is taken care of by rtreating two letters – typically I and J, less frequently U and V – as equal.

Apparently, the present grid combines nine more letters to equal pairs. The hint says that the Playfair grid goes from left to right and from top to bottom. There doesn't seem to be a keyword, so one possibility could be:

possible letter pattern

What are the numbers around the perimeter for?

The OP has disclosed that the answer is a word. There are thirteen numbers. The Playfair cipher works with pairs of letters. If there is an odd number of letters, there must be padding. This doesn't really fit, but perhaps "Playfair" doesn't refer to the Playfair cipher, only to the practice of pairing letters.

Continuing the pattern above, so that all of the thirteen numbers are assigned a cell, we get:

17267 → 14: U or V
2008 → 12: S
18877 → 3: D or E
20793 → 5: H or I
16078 → 9: N
15177 → 13: T
5254 → 5: H or I
20689 → 9: N
21959 → 12: S
5046 → 13: T
18786 → 3: D or E
12321 → 1: A
16637 → 3: D or E

The numbers after the arrows are the cell numbers, starting with 1, 2, 3, and 4 from left to right in the top row and ending with 16 in the bottom right. The letters correspond to the scheme I've posted above, which isn't established at all. That result doesn't look promising. There is a good share of vowels, however, so maybe we should build an anagram? Nah, that would be devious.

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  • $\begingroup$ To add to the pattern with 0 in the bottom corner, each row and each column contains 2 odds and 2 evens, in the rows the sum of the evens is 2 less than the sum of the odds $\endgroup$ – lPlant Oct 4 '17 at 16:33
  • $\begingroup$ With the premise that the grid goes 0-15 instead of 1-16, then when solved as a sliding puzzle the blank should be in the upper left corner, which is solvable, unlike in my previous comment. May or may not be relevant. $\endgroup$ – Bobson Oct 4 '17 at 17:39
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    $\begingroup$ 31 would be logical as the last number. No idea what to do with that, though. $\endgroup$ – Hugh Meyers Oct 5 '17 at 6:40
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    $\begingroup$ @HughMeyers: Yes, with the pattern that AstroMax found I get 31, too. When I get there, every other cell has exactly two numbers. $\endgroup$ – M Oehm Oct 5 '17 at 16:55
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    $\begingroup$ "That result doesn't look promising" Doesn't your result spell "USE INT INSTEAD"? No idea what that's supposed to mean but it does look promising to me. $\endgroup$ – SirGrapefruit Oct 6 '17 at 16:19
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Partial Answer

Using the hint given which suggests there is three parts to the puzzle, a potential answer to the one part may be.

24. In the box if you start at #1 counting in order (ie 1,2,3,4,5 etc.) towards the right, then continuing counting on the rows below, a pattern seems to emerge. Starting at 1 the number 2 is two boxes away (not counting the box you start on). If you count from 2 to 3 the number 3 three is three boxes away. From 3 to 4 four boxes away and so on. This pattern continues all the way up to 15.

Hence

If we follow this pattern counting in order, the amount of boxes away we are from the previous numbered box, we could conclude that the blank box is 8 boxes away. But as the OP commented the number is higher than 16. If the numbers cannot be repeated then you may guess 24. if you count past the blank box as to not repeat the number 8 you would get 24. As M Oehm has uncovered, 31 is much better suited to complete the pattern rather than 24.

The surrounding numbers

could possibly be decrypted into a key. The OP mentions playfair in previous comments so a key would prove of useful for a playfair type encryption.

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  • $\begingroup$ that's a good reason to place 0 there $\endgroup$ – JMP Oct 4 '17 at 20:45
  • $\begingroup$ You're on the right track, you found the pattern (although your analysis of it is wrong), so we have a pattern, a playfair square, and some huge numbers. What do you think you should do? $\endgroup$ – Quadraxus Oct 5 '17 at 5:59
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Part 1 Part 2 Uh I'm sorry guys. All I wanted was advice I guess I should just not bother with making puzzles. I never found an audience and everyone seems to hate them....

You separated the 3 components and eventually found USEINTINSTEAD, what you then had to do was use the INT() function. MOD uses the remainder of a quotient, and INT uses the integer part of the quotient. Do that and you get 539, 62, 589, 649, 502, 474, 164, 646, 686, 157, 587, 385, 519. INT those by 32 and we get 16 1 18 20 15 14 5 20 21 4 18 12 16. Translate this via A=1 B=2.... to get PART ONE TUDRLP. MOD those by 32 and translate them same way you did and we get O/P A Q/R T T W/X O/P H/I N D/E B/C A S. You may have to do a little bit of guess work but you can see that it starts off with PART TWO H/I N D/E B/C A S, you may have to do a little bit of experimentation but given what you know you will find out it is a brick wall for thunderclaps. IE put part one on top of part two and read down left to right.

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    $\begingroup$ Why do you think people hate them? $\endgroup$ – MildlyMilquetoast Oct 8 '17 at 4:42
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    $\begingroup$ I can't read your handwriting - would you mind typing up a clearer explanation? $\endgroup$ – Deusovi Oct 8 '17 at 5:02
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    $\begingroup$ @Quadraxus "It would be a shame to just let this peter out, because I think there are some good concepts in the puzzle." "This puzzle looks nice!" and a score of 7. People don't hate your puzzles. Just because a puzzle is hard or unintuitive in places doesn't mean that people don't like it. Once you know the answer to a puzzle, it's hard to gauge the difficulty of it. Don't feel bad. $\endgroup$ – MildlyMilquetoast Oct 8 '17 at 5:06
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    $\begingroup$ don't be disheartened if some of your efforts seem to not succeed - justbereal.co.uk/… $\endgroup$ – JMP Oct 8 '17 at 8:35
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    $\begingroup$ So yeah I put this in puzzle design because I thought this was obvious (obvious yet difficult), I felt like you didn't really have to make any leaps in logic but then again I set the puzzle and I have Asperger's so I think differently. Oh uh sorry I shouldn't have said that should I..... $\endgroup$ – Quadraxus Oct 8 '17 at 16:21
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The answer is

16

Reasoning is

the box has the numbers 1-15 without repeating. 16 would complete the box while keeping the sequential pattern. The numbers surrounding the box and the lines inside seem to be just a distraction from the solution.

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  • $\begingroup$ No, the other components are obviously a part of the puzzle as well. So no. Also I will give hints if necessary. $\endgroup$ – Quadraxus Sep 28 '17 at 11:49

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