CTF: BITSCTF 2017
\nPoints: 20
\nCategory: Crypto<\/p>\n
\nBrute and get the base 32 format of flag.
\nencrypted.txt:MZYVMIWLGBL7CIJOGJQVOA3IN5BLYC3NHI<\/code>\n<\/p><\/blockquote>\nThis task is worth 20 points, but only 8 teams have solved it during ctf and I really wonder why.<\/p>\n
Before we start, I assume that everyone knows how base32 works: link<\/a>.<\/p>\n
Solution<\/h2>\n
Task description tells us that flag is converted to base32 and somehow encrypted.<\/p>\n
From other tasks we know flag format and when we compare it with ciphertext length, we can assume that plaintext looks like this:
BITSCTF{*************}<\/code>.<\/p>\nLet’s encode first five letters of flag (one block of base32),
BITSC<\/code> to base32:IJEVIU2D<\/code>.<\/p>\nCompare first 5 letters of base32 plaintext and ciphertext:<\/p>\n
\r\nI J E V I U 2 D\r\nM Z Y V M I W L\r\n<\/pre>\nWe can notice that every letter in ciphertext decodes to distinct letter in plaintext (with
M<\/code> decoding twice toI<\/code>), so we can guess that this is kind of monoalphabetic substitution cipher.<\/p>\nLet’s look for any patterns in ciphertext alphabet.<\/p>\n
Our alphabet(all base32 letters):<\/p>\n
\r\n A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 2 3 4 5 6 7\r\n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31<\/pre>\nEncrypting alphabet<\/p>\n
\r\n A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 2 3 4 5 6 7\r\n ? ? ? L Y ? ? ? M ? ? ? ? ? ? ? ? ? ? ? I V ? ? ? J W ? ? ? ? ?\r\n 11 24 12 8 21 9 22 <\/pre>\nSo:<\/p>\n
\r\n3 -> 11\r\n4 -> 24\r\n8 -> 12\r\n20 -> 8\r\n21 -> 21\r\n25 -> 9\r\n26 -> 22\r\n<\/pre>\nWhen we look closely we can see that this is encrypted with affine cipher<\/a>, with
a = 13<\/code> andb = 4<\/code>.<\/p>\nBy the way, after finding that pattern we realize that title of this task is anagram of affine<\/b> word.<\/p>\n
So our encrypted alphabet will look like this:<\/p>\n
\r\n A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 2 3 4 5 6 7\r\n 4 17 30 11 24 5 18 31 12 25 6 19 0 13 26 7 20 1 14 27 8 21 2 15 28 9 22 3 16 29 10 23\r\n E R 6 L Y F S 7 M Z G T A N 2 H U B O 3 I V C P 4 J W D Q 5 K X\r\n<\/pre>\nNow we can get our plaintext:<\/p>\n
\r\nMZYVMIWLGBL7CIJOGJQVOA3IN5BLYC3NHI -> IJEVIU2DKRDHWUZSKZ4VSMTUN5RDEWTNPU\r\n<\/pre>\nAfter that we need to add b32 padding and finally we can read our solution:<\/p>\n
\r\nimport base64\r\nbase64.b32decode('IJEVIU2DKRDHWUZSKZ4VSMTUN5RDEWTNPU======')\r\n-> BITSCTF{S2VyY2tob2Zm}\r\n<\/pre>\nEaster egg:
\nflag is base64 of string ‘Kerckhoff’:<\/p>\n\r\n$ echo -n \"S2VyY2tob2Zm\" | base64 -d\r\nKerckhoff\r\n<\/pre>\nKerckhoffs’s principle<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"
CTF: BITSCTF 2017 Points: 20 Category: Crypto Description Brute and get the base 32 format of flag. encrypted.txt: MZYVMIWLGBL7CIJOGJQVOA3IN5BLYC3NHI This task is worth 20 points, but only 8 teams have solved it during ctf and I really wonder why. Before…<\/span> <\/p>\n
Read more ›<\/div>\n<\/a><\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[26,25],"tags":[20],"yoast_head":"\n\n\n\n\n\n\n\n\n\n\n\n\t\n