CRACKME RE/Math challenge MEDIUM (Decrypt the hidden message)

re
crackme
cracking

(Diaze Haze) #1

Something I came up with after the CTF, some of you may have a clue how to do it, there is more than one solution!

However I will point out now this will test almost all you’re skills is logic to do properly (there is of course the non proper way)

The only chars in use are in the ASCII range 0-127 or 1-128, however you prefer it.

If no one gets anything at ll I will post a few hints slowly over the upcoming week w/s 17th dec

216 202 216 64 224 217 133 227 146 149 223 215 143 216 64 234 206 146 229 140 213 230 142 222 143 218 137 154 216 137 220 231 140 209 218 194 137 223 133 215 212 135 222 216 217 136 147 224 212 232 203 202 224 200 136 150 81 153 162 87 147 213 221 208 64 202 221 215 226 202 202 226 215 221 202 64 208 221 213 147 87 162 153 81 150 136 200 224 202 203 232 212 224 147 136 217 216 222 135 212 215 133 223 137 194 218 209 140 231 220 137 216 154 137 218 143 222 142 230 213 140 229 146 206 234 64 216 143 215 223 149 146 227 133 217 224 64 216 202 216

To check you solved it yourself, sha256 the answer then md5 it, the result should be

10c438bf7b9fa3746238df966b712360

Example given:

$ sha256 -s test_string
SHA256 (“test_string”) = 4b641e9a923d1ea57e18fe41dcb543e2c4005c41ff210864a710b0fbb2654c11
$ md5 -s 4b641e9a923d1ea57e18fe41dcb543e2c4005c41ff210864a710b0fbb2654c11
MD5 (“4b641e9a923d1ea57e18fe41dcb543e2c4005c41ff210864a710b0fbb2654c11”) = 3babffc2e6e579831fe64e9b4ddb4cec

In this example the hashcheck you have the right password/string is 3babffc2e6e579831fe64e9b4ddb4cec

For those of you that do manage it!

Post a single character in this form:

block[n] = c

where n is 0 to 139


(Diaze Haze) #2

Just to be nice because this is a little meaner than it looks and was pretty much made to break freq analysis:

block[138] = e


(bin shift) #3

Sounds like a great challenge. For those of use who are new to hash reverse engineering, please publish the steps on how one would go about reverse engineering. I have only played around with simple substitution ciphers, but never have attempted to crack a hash.

Thanks

Bin


(Diaze Haze) #4

Really no one? Next clue! a big one

block[3] = space
or \s or ’ ’


(Diaze Haze) #5

ok dokey a bigger clue perhaps, look at the first 4 and last 4 blocks:

First(4) 216 202 216 64
Last(4) 64 216 202 216

whistles


(Diaze Haze) #6

The final tip :slight_smile:

[19:37:53] ANON> plaintext starts with lel by any chance?
[19:38:14] diaze> nope and post in forum :slight_smile:
[19:38:29] ANON> eh then i got it wrong
[19:39:18] diaze> its deceptively easy
[19:39:37] ANON> i got that it’s palindrome
[19:40:03] diaze> its not


(Diaze Haze) #7

Final hint -_-

First word/chars:

blk[0,1,2,3] = [216,202,216,64] = [s,e,e, ]