Incredibly tough challenges from competitors

[Madness]

If it wasn’t “incredibly tough” enough, here is one with randomized key grid:

JQQJYJ A4OJ5F NDUAPF OCQGEF HOKVXC REOAXH E3RM5P CFRQG3 JZ2PT3 RZIDPT
QGNNUR JJDOXH RSDLS3 WJ5FTU D23QG3 PQZS3U TWIDPT DGKYY3 JCJDOJ AP3URT
QGKC5V CNSCHT NJY2IW C3Y3Y3 GURAAC RCEA5V 5NSKVQ YFZJTU H3XCGT DGNS3U
TWJINP T5IO5V Y3GRMA OMU5NJ AAHA5V CEFQG3 RAHSYJ ZS25ZF NCXCHM OYFKNF
GYUXCG TSOQG3 JYQZJT CAFUDP KX23RF RSDPYM HC5EAT YFY3UX RSFHFT DHXCKP
OMRQG3 GR54AC QT2XCG LD55QG BHEA5F QYH2LJ Y3ULAC 5AC3TT NXCKOJ YEA5FT
QG2WD3 AROAXH D3RFXU R3DXVD JILJ5V EOUUSF O2ZTMU GXSOMA YAY2LJ YEAQG3
TNQA5J QYCQG3 YZMLPF WMHQCV ZX23RF 35JWAA YKCWJU BHAQG3 RNQA5J QXKCWP
5FRQEP OMSTDP BHHWY3 KSMLMT JJY2JT 5VUTXU KJWFFR TUTI3U WJLRX3 CHKGRU
LNCTRF D2EYVM TPJDOJ GSZRTF BMQG3R ZIGHXF OHD3Y3 YUSP5V 53YHX3 GY3QRS
C2OAEO OHHWNQ TA5J5V RULP3U RRQGKY GYNSKC RWPYNJ EHZKHA NQYCHT ZPDODT
QG3GRU NMDUZJ L23RFX NLJU5F HCQGNP TDYCWJ Y3UWJX EAY3XC TGUTPQ ZPICRW
YEAUTT QG2WD3 LRDMT3 PXCGFH TMD2MQ LVUDFQ 35XAHA WFECQQ BCKV3R 3MJTRU
OCCCAC CPDENF JFFFRT GMSIKM BHD3RT QGNXHR BMJXEO QOAIOD LAEFYU NQZIDP
YCPNSV 53RGXU HENHWU HHD3Y3 YAQAWU GYAQG3 BHKTSF HZ3OZF GYCHXU NLXYJT
D3YMCV NX2S3U GA23DP WIOZEF HHHG3R CCPNSS RPQCVT YKVXCG LDLIG3 LKAMCV
GJ5QCV RQDP5V CESDAW OHCPEA RNQATW E3RVMA QSYC5V 5NPYNJ QHOANY RPJU3T
EGKYY3 NCJDOJ BXDH3T DG3XUT R53RIF QA2EAU ZIGCLY RI2LJU OHPPPQ CKHAI3
PLXYIW EGKCGR RFN2YQ TZJWJU DSG3ZJ T5KUDY EYCQG3 OYXUWP CFPD5F GYCN2Q
ZVHG3R JPYHCO

[Upsidedown]

Yes, I tried a couple of other techniques but a dictionary attack was the way to go.

[Madness]

Did you try looking for assignments of digits to letters so that all blocks could be multiples of 7?
I’m trying that now, with some tricks to shorten the search.

[Upsidedown]

Yes, I also noticed that the last letter in every block is encrypted with the same monoalphabetic substitution and that the plaintext and ciphertext of a letter must be on the same row of the key square, so by combining multiple plain/cipher pairs you can determine that a set of characters must be on the same row.

If you perform frequency analysis on the last item of each block it gives you a good guess of which characters are on each row, and if a letter is encrypted as itself then it must start the row. Combined with the fact that the lower number must be a multiple of 7, I think that should significantly reduce the search space.

[Falsethunder]

That was fun. I ended up just generating one HUGE vignere key and then solving that directly.

[The-letter-wriggler]

VIGENERE 2 KEYS
I can’t get the TWO keys but this ONE key solves it.
A 30 letter key + TLW (all in capitals of course)

MD5: c3cef17eb1c43fe08961a8c96ceec2b6

If you do not have this key to check it
I’ll post the actual key itself if you want it.

I have no idea with the 3 keys Vigenere.

[Madness]

@TLW keep at it. The keywords for each challenge are coprime. That should help.
(Your MD5 does not match for number 2. Even one letter off can cause that to happen.)

[Author]

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