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So Bob and his friend were now enemies. They were talking about how annoying the other one was when Bob discovered a note his friend had written. BTW Bob's friend is a male, he couldn't decode the message, but maybe you can. He has not done geometry yet.

Decode this:

72.14.80.17.20.54.8.114.95.24.65.48.54.98.126.9

Helpful Hint:

Pie out of a stick.

Clue:

Bob is in Geometry right now, and working with circles.

Also, please comment why you downvote it if you downvote it. Otherwise, I will not know why you downvoted it, therefore cannot improve.

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    $\begingroup$ Not downvoting, at the moment, but one small bit of advice: If the puzzle cannot be solved, without the hint, then the hint may need to be part of the puzzle, elsewhere. :) Consider a hint to be something to help people struggling with the puzzle, and the clues to be required information. $\endgroup$ May 3, 2016 at 13:13
  • $\begingroup$ +Khale_Kitha thanks for the bit of advice $\endgroup$
    – TigerGold
    May 4, 2016 at 0:26

2 Answers 2

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My answer is

youregettinggood

reasoning:

From the hint I guessed that it just meant looking at the digits of pi. I then lined those digits up with the digits that were provided in the puzzle. Curiously, each of the digits of pi divided the corresponding integer in the sequence 72.14.80.17.20.54.8.114.95.24.65.48.54.98.126.9. That is, 3 | 72, 1 | 14, 4 | 80, and so forth until the end of the provided sequence.

Performing this division gives the following numerical sequence:

24 14 20 17 4 6 4 19 19 8 13 6 6 14 14 3

Turning these digits to letters (A = 1, B = 2, ...) this now becomes:

xntqdfdsshmffnnc

and shifting the alphabet over 1 letter yields the answer "youregettinggood"

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    $\begingroup$ If you consider A=0, B=1... then you don't need the shift. Nice answer, btw! $\endgroup$
    – ffao
    May 5, 2016 at 5:04
  • $\begingroup$ Thanks! That's true, indexing by zero would have simplified it! By the way, do you by any chance compete on Codeforces? Your username seems familiar and I think I've read some posts from you on there! $\endgroup$
    – Andr0s
    May 5, 2016 at 5:12
  • $\begingroup$ Yep, I do! Makes sense that you've seen one of my comments, over there I'm a bit too talkative and outspoken for my own good... $\endgroup$
    – ffao
    May 5, 2016 at 7:46
  • $\begingroup$ Very good answer. +1 $\endgroup$
    – hexomino
    May 5, 2016 at 8:59
  • $\begingroup$ You are correct! $\endgroup$
    – TigerGold
    May 7, 2016 at 16:59
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My guess is

WANTS PLAYER TWO

Firstly, the hint

I think refers to the fact that you consider each number in the sequence modulo some fixed number representing the length of a pre-defined alphabet. The number line (stick) is forced into a cyclic pattern (pie) via modulo arithmetic.

In this example, I chose my alphabet to include the letters A to Z and a 'space' making it of length 27. Considering each of the given numbers modulo 27 we get

18.14.26.17.20.0.8.6.14.24.11.21.0.17.18.9

Then

With the resulting number sequence, I chose 0 to represent 'space', thereby getting three distinct words:

18.14.26.17.20    8.6.14.24.11.21     17.18.9

Then, I chose a mapping from the set of numbers 1-26 to letters of the alphabet A-Z, a substitution cipher, with the relevant numbers as follows:

6 $\rightarrow$ L
8 $\rightarrow$ P
9 $\rightarrow$ O
11 $\rightarrow$ E
14 $\rightarrow$ A
17 $\rightarrow$ T
18 $\rightarrow$ W
20 $\rightarrow$ S
21 $\rightarrow$ R
24 $\rightarrow$ Y
26 $\rightarrow$ N

which transforms the words into WANTS PLAYER TWO

Context

In most 'classic' puzzles referencing Bob, his friend is Alice and they are usually playing some kind of idiosyncratic game. Alice is nominally player one and Bob is player two. It would make sense that, if they had some sort of falling out, Alice would miss Bob and his willingness to participate in new things. Thus the reason she 'wants player two' again.

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  • $\begingroup$ Not quite. Although you're kindof close with the shift, not quite on. $\endgroup$
    – TigerGold
    May 4, 2016 at 0:22
  • $\begingroup$ By the way this is very clever, you got +1! $\endgroup$
    – TigerGold
    May 4, 2016 at 20:10

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