Error 115

[Towerfbabble]

Error 115 Error 115Error 115vvError 115Error 115Error 115vError 115Error 115vvError 115Error 115Error 115vError 115Error 115Error 115vError 115Error 115vvError 115Error 115Error 115Error 115Error 115vError 115

[CptArel716]

How do you know thats wrong?

[Towerfbabble]

How do you know thats wrong?

Because I did it...

[CptArel716]

What im getting at is the letters are encoded you need a Polybius Square to decode the letters. read this link

http://www.artofproblemsolving.com/blog/27171

so any 6 letter word that we have come up with could be right.

[Towerfbabble]

What im getting at is the letters are encoded you need a Polybius Square to decode the letters. read this link

http://www.artofproblemsolving.com/blog/27171

so any 6 letter word that we have come up with could be right.

I did that.

[CptArel716]

What im getting at is the letters are encoded you need a Polybius Square to decode the letters. read this link

http://www.artofproblemsolving.com/blog/27171

so any 6 letter word that we have come up with could be right.

I did that.... You can't make those into 5x5. If we have to we would need 2.3 keywords.

here is an example

_ A D F G X

A a b c d e

D f g h j/i k

F l m n o p

G q r s t u

X v w x y z

So looking at this we can see that the letter pair AA=a AD=b AF=c and so on this is what I'm talking about.

So with your key word "portal" the encryption would look like this, XG GD FF AF GF GF DD GF GG XX FF FF FG FG FD FG FG FD GG FF FD GF AA FG XF DX FX DX GF XF GG FF FA -- -- AF.

Now with this 5x5 Polybius square it would be, NBNSDONNTGXWTLRSYJNJSOXNNXNRLEOXUS which is just random letters but don't count any words out yet, unless we have undeniable proof that we know the key word.

[Towerfbabble]

What im getting at is the letters are encoded you need a Polybius Square to decode the letters. read this link

http://www.artofproblemsolving.com/blog/27171

so any 6 letter word that we have come up with could be right.

I did that.... You can't make those into 5x5. If we have to we would need 2.3 keywords.

here is an example

_ A D F G X

A a b c d e

D f g h j/i k

F l m n o p

G q r s t u

X v w x y z

So looking at this we can see that the letter pair AA=a AD=b AF=c and so on this is what I'm talking about.

So with your key word "portal" the encryption would look like this, XG GD FF AF GF GF DD GF GG XX FF FF FG FG FD FG FG FD GG FF FD GF AA FG XF DX FX DX GF XF GG FF FA -- -- AF.

Now with this 5x5 Polybius square it would be, NBNSDONNTGXWTLRSYJNJSOXNNXNRLEOXUS which is just random letters but don't count any words out yet, unless we have undeniable proof that we know the key word.

XG= y

You did t wrong

6x6: XG=0

[Flammenwerfer]

Well, I just tried "zombie" which was just gibberish.

[CptArel716]

Where do you get 6x6 from this is an ADFGX cipher if the cipher text has only the letters A, D, F, G, and X, then it is an ADFGX cipher. It is not an ADFGVX cipher because there is no V in the text.

[Z0M3IE_SLAYA_115]

maybe the keypord is death

[ThrowAmonkeyBomb]

what is this? :lol: