Old stuff, but heh too lazy to update my blog.
Thanks to Radek Suski because entertain my report :D
Easy crypto: 82 73 81 81 61 83 52 62 63 41 74 22 01 42 73 31 74 52 01 21 74 01 71 83 83 01 92 73 83 32 61 73 01 81 62 43 01 74 63 51 01 71 43 92 32 92 92 22 23 43 42 61 01 51 51 72 41 62 81 72 42 22 91 01 74 91 52 74 01 42 33 23 61 23 43 21 61 52 22 01 52 74 42 01 82 63 43 94 92 93 91 32 23 32 01 94 43 01 21 62 21 74 52 81 32 71 91 33 93 01 43 51 63 42 01 32 01 52 53 32 94 63 74 91 42 01 52 81 83 92 92 01 21 74 01 91 31 41 53 82 91 62 63 01 73 42 82 91 21 22 51 71 01 91 94 32 01 51 83 22 62 61 51 63 52 94 82 01 41 33 01 83 73 01 92 82 41 62 61 41 32 42 83 01 61 94 91 53 42 52 21 01 41 93 53 82 91 31 01 63 53 92 01 52 43 93 23 82 81 72 93 01 81 94 41 21 82 61 61 43 83 01 93 42 91 31 94 42 01 61 83 01 83 21 72 33 41 42 51 01 81 23 43 01 42 93 61 23 91 32 53 63 01 21 62 21 74 52 81 32 71 91 33 93 01 73 33 53 81 01 21 22 21 53 01 32 22 52 71 72 83 32 93 01 83 92 51 61 53 91 83 93 91 52 01 81 23 43 52 53 33 82 01 31 43 71 33 31 82 93 61 41 62 01 93 51 52 73 93 91 91 31 01 52 43 52 94 74 51 41 01 52 94 01 41 41 01 81 23 43 01 53 42 72 21 01 41 94 93 33 91 01 91 51 83 32 52 01 21 91 33 01 33 93 42 01 74 62 93 01 92 21 53 01 32 94 51 83 93 91 01 41 33 01 63 53 92 01 51 74 43 31 53 32 42 51 01 81 23 43 01 33 53 92 31 23 82 31 01 92 52 32 93 01 92 42 01 51 32 62 33 71 01 94 92 53 01 23 61 23 43 21 74 42 23 92 51 01 52 74 83 94 01 33 92 51 74 43 74 01 94 93 33 51 32 83 91 22 82 52 42 61 01 81 94 71 42 53 94 22 01 81 82 42 01 32 42 51 01 53 74 42 53 31 41 81 21 61 52 22 01 31 94 82 91 01 91 31 42 92 94 81 62 74 73 63 ------------- Description: ------------ Look carefully! Does it look like hex?the numbers refers phone keypad. decode it and then you'll get another cipher which is a vignere cipher. decrypt it and you'll get an article about cryptography. the 'key' is a hint to this challenge's flag.
FN1hJU9XAJZhTF8qbnJENt1XBMd6i0utzJpHCGyVt4yp8LsHYHUJP+/M+37eNjldkx4T5xnliSrIsz/qQHB9PA==as you can see,this is just a normal base64. decode it and you'll get some scrabble/rubbish/random words+symbols.view hex of the decoded base64 just like below.

ndeunf ofqugzdg cdru kuus zdpul dsl giyisp uatvsuseiannf ev kgisp lvzs ovgghtecl pvrugswusey dsl eh ogudeu egdsytdgusof dsl kgispisp tvzug kdom ev ecu tuvtnu!zlugu ecugu iy sv riyivs,ecu tuvtnu zinn tugiyc svz dsyzug;zcioc pgvht zdy gfds onudgf doohyul vx ev ku dyyvoideul ziec?wl5 ecudsyzug dsl yhkwie ecu yhw.ecde iy fvhgxnp.End? not yet! you still need to decode the 2nd cipher,Substitution Cipher! Here's a wikilink for this cipher. how this cipher works? here's an example rabbit - real words. so the person who want to send this message want to cipher it,so he substitute letter r = x, a = w, b = e, i = k, t = u so the rabbit,once ciphered will become xweeku. so,how to get the real answer for this challenge? reverse the decoded words up there :P thanks.
fvrwslwslswhgacsremfberbubgrihgbtvi
#!/usr/bin/python
charset = 'abcdefghijklmnopqrstuvwxyz'
encoded = 'fvrwslwslswhgacsremfberbubgrihgbtvi'
keys = open("wordlist.txt", "r")
for key in keys.read().split('\n'):
message = ''
for i in range(len(encoded)) :
p = charset.index(encoded[i])
k = charset.index(key[i % len(key)])
if k - p < 0 :
message += charset[((p - k) + 26) % len(charset)]
else :
message += charset[(p - k) % len(charset)]
if message[:-2].count('the')>1:
print 'The key: ' + key
print 'The message: ' + message
break
OHMAIGAWD! It seems like we've been hacked! But what did the hackers steal? From the logs, it seems like they exploited an SQL injection bug on our website. Help us find the name of the database that they stole and we shall reward you handsomely.So players were given with a log to analyze them. You can get it here