Challenge 7B

[elsueon]

Are there any hints for challenge 7B? I have done the first part but can’t get anywhere with the second, even using the type of cipher and way of reading that’s mentioned.

[f6exb_the_frenchy]

@alsueon
What is the lenght of the cipher text ?
What are the possible divisors ?

[sofiabunny]

I have the right keyword but still cant get past the next part as the text still doesn’t make sense. any more tips Harry? thank you

[f6exb_the_frenchy]

@elsueon (2)
When you have cut your text in columns, take the first letter of each columns and put them in an in line anagram solver.

[sdv_x]

idk what Harry and the Elves will let through but:

try searching up ciphers (on wikipedia) which frequency matches that of english
divide text into chunks. number of chunks = key length [this will probably be removed sorry]
use text editor (e.g. onenote) to move chunks of text around until plaintext emerges

extra hint that probably will also be removed – often, keys have been part of the title

[elsueon]

Length is 560 characters (I think??) and I have checked all factors/divisors. Still can’t get anything that makes any sense!

[f6exb_the_frenchy]

@elsueon
If I choose 7 as factor (don’t ask me why), and if I cut my cipher text in 7 columns, I have (copy-paste of my program):

colonne 0 AIYNAGKRCTELUOISGAEELLSONOKONSNAAOTOWDSMSEAATOEPNCEDTLVUNTRLIDDTHHEPTEFAIADGOSWR
colonne 1 EMUIREINMOUSRSGSIEHNYARSRESNORROEDRELEONPESEEHUAEMAIOCRBRMEIISFELTNIRTEYLVAOTTTE
colonne 2 SROTRNWAOIQIIKNEANTOAETNTTIEOETNTFUNIDFEETMSRTMHEOCDIEEAYECYFIOHESAPECNRLDWDNSII
colonne 3 LNTGEIOAADSEEBSEDOUASLITRTESROSOSTALEEIAYAGLNYOESOEAROOAMEOSNISGAIAHALNSTECITTEU
colonne 4 CDRSIWLGMNITEMPENNEDWVUFYLLSNVYRDEEELTRTALODPISSDPRSNTYLASEDEPTRPODNCICBNARNAYHS
colonne 5 POAALVYLEOARRDILNSSYIIECOADEUEETEAPADSAPBREPOLSTIFLECCIVERSAEDOOCWRSGOOEUENNOCLQ
colonne 6 EFEESNRRCARDQOSAACRRPLNIEHSTEBEMTERSDNLYATREOTNMERTNELNLAIUITFEENTGIFLISWDENCMSE

Now, take the first letters of each column and search in an anagram solver wich words it gives.
It is also possible to test the second or third letters to polish your answer.

[The_Letter_Wriggler]

FOR 7B
HOW TO USE THE FACTORS
======================
The Factors :
1 2 4 5 7 8 10 14 16 20 <|> 28 35 40 56 70 80 112 140 280 560

Pair the factors like so:

1 560
2 280
4 140
5 112
7 80
8 70
10 56
14 40
16 35
20 28

The first number is the keyword length (that you decide to use)
the second number is the length of each row.
So for a 7 letter keyword we need 80 letters per each row

[I_am_Sid]

If Harry allows it the key length is 7

[The_Letter_Wriggler]

Sorry, in my last post I used the word factors it should be divisors.

Sometime us crackers are, well, just crackers!

[Author]

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