15
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Since everyone is trying their hand at this type of puzzle, I've decided to create a number seq -

We interrupt your Puzzle Solving Entertainment to inform you of this Public Service Exclamation:
This is NOT a number sequence puzzle, although you will be finding a number from a series of numbers.
If you're looking down upon this sequence, I'll tell you that it's a very basic puzzle.

-uence looking puzzle that isn't actually a number sequence.

565, 721, 575, 425, 667 => 4??

(EDIT: The fourth number should have been 4TWO5 not 415!!! Super sorry for the inconvenience)

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  • 2
    $\begingroup$ The italicized letters are i or o. With "basic" I'm guessing binary, and the italicized letters become 101,110,101. $\endgroup$
    – MikeQ
    Jan 19, 2017 at 6:49
  • 2
    $\begingroup$ Which makes 565, the first number of the sequence... $\endgroup$
    – boboquack
    Jan 19, 2017 at 6:52
  • $\begingroup$ Or, if we do not separate them, 373. $\endgroup$ Jan 19, 2017 at 6:54
  • $\begingroup$ I suspect octal numbers since no digit is over 7. $\endgroup$ Jan 19, 2017 at 6:55
  • $\begingroup$ Converting from octal to decimal, the sequence is 373, 465, 381, 269, 439. No clue what it means. Maybe the "NOT" is relevant? $\endgroup$
    – MikeQ
    Jan 19, 2017 at 7:10

1 Answer 1

8
$\begingroup$

The number is

416


As @MikeQ and @boboquack noted in the comments,

The italicised letters are ioi iio ioi, which, when taking "basic" aka hint to binary", leads to 565, the first number in the "sequence". In addition, each line in the blockquote has three italicised letters, confirming this interpretation.

If we then do this for each number, we get

101 110 101
111 010 001
101 111 101
100 010 101
110 110 111
formatting according to the "looking down upon this sequence" hint.

Now, we want three digits, and our grid is neatly split into three blocks, but none look particularly like a digit. However, with some nudging from the OP,

we can re-group the bits so that all the fours bits are together, all the twos bits are together, and all the ones bits are together. For example, looking at the first row, the fours bits from left to right are 1__ 1__ 1__, the twos bits are _0_ _1_ _0_, and the ones bits are __1 __0 __1. We can put them together to form 111 010 101.

Doing so for each row yields

111 010 101
100 110 101
111 010 111
101 010 001
111 111 001

If you can't see it yet:

*** * * *
* ** * *
*** * ***
* * * *
*** *** *

That doesn't start with a 4 though.

Let's flip it. (Or, more technically, take the bits in little endian order).
* * * ***
* * ** *
*** * ***
..* * * *
..* *** ***

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  • $\begingroup$ I've been looking at this for ten minutes and still have no idea what you mean by 'We can re-group the bits so that all the fours bits are together, all the twos bits are together, and all the ones bits are together.' It would be great if you could be less vague $\endgroup$
    – Wen1now
    Jan 21, 2017 at 3:40
  • $\begingroup$ @Wen1now I have edited that part, hopefully it's clearer now. $\endgroup$
    – Volatility
    Jan 21, 2017 at 3:49
  • $\begingroup$ @Wen1now in other words, look at the shape the odd digits make, then look at the shape the 2+ mod 4 digits make, then look at the shape the 4+ mod 8 digits make $\endgroup$ Jan 21, 2017 at 4:03

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