4
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3 2 1 Go

Here the key has already been viewed, be fooled you will, taken from the left to the right it shall be revealed!

ily all sha llh aou

Hint:

{3} to 13th

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  • 1
    $\begingroup$ Provide an explanation for why you downvoted this 9 10 11 12 13, if you may of course $\endgroup$ – Kyle Feb 29 '16 at 13:47
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    $\begingroup$ Yoda I see...... $\endgroup$ – ABcDexter Feb 29 '16 at 13:51
5
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It's:

You shall all hail
Because I can read it scrambled, but don't see how "3 2 1 Go" makes it so.

EDIT: Or is it, thanks to @manshu and @ABcDexter

Shall hail all you.
Because that starts in the 3rd group grabbing "sha llh a" which needs "il".
The gets "all" from the second group and finally "you" from the first.

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  • $\begingroup$ Almost there... $\endgroup$ – Kyle Feb 29 '16 at 13:54
  • $\begingroup$ Words were made, but sense they don't make $\endgroup$ – Kyle Feb 29 '16 at 13:56
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    $\begingroup$ @Mr.Derpinthoughton "Shall hail you all..." ? $\endgroup$ – ABcDexter Feb 29 '16 at 14:02
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    $\begingroup$ "All shall hail you"? $\endgroup$ – Ian MacDonald Feb 29 '16 at 18:54
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    $\begingroup$ @IanMacDonald Only $4! - 8 = 16$ iterations to go :) ) $\endgroup$ – Paul Evans Feb 29 '16 at 20:14
3
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Explanation:

"Can you do this?" has 13 visible characters
3 2 1 go => the number 3 is right above "Here"
Here...

The sentence was divided into 5 pieces made of three letters each

ily all sha llh aou

Until this point a little bit of guessing is needed

Hint:

{3} to 13th
{ily} all sha llh.... a... ou
.........................13th

{3} to 13th means a group {} containing 3 letters to the 13th thirteenth position

all sha llh a{ily} ou

Thus:

all shall hail you

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  • 2
    $\begingroup$ sigh beat me to it. $\endgroup$ – Hugh Meyers Feb 29 '16 at 14:15
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    $\begingroup$ Ah, we would have got in at most $4!$ tries :) ) $\endgroup$ – Paul Evans Feb 29 '16 at 14:16
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    $\begingroup$ @Deusovi There are 4 words. You can create all possible iterations of 4 things in $4! = 24$ ways $\endgroup$ – Paul Evans Feb 29 '16 at 20:22
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    $\begingroup$ @HughMeyers This answer was provided by the person that posed the puzzle in the first place. I wouldn't really consider that a "beat me to it" scenario. :) $\endgroup$ – Ian MacDonald Feb 29 '16 at 20:41
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    $\begingroup$ @IanMacDonald Me neither, the author is certainly a cheater in my point of view :) $\endgroup$ – Kyle Feb 29 '16 at 20:43

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