We searched for "iip" and knew that it's a name of The Invisible Internet Project (I2P), so we installed it and tried to do some serfing, all're OK. But what's the strange link in the task? While we serf i2p we know, that every site in i2p has usual addres in domen .i2p that's usual alias from long hash. Every such hash is over 500 characters and it must end in "AAAA" in i2p router we can add this hash as our i2p site and router would generate us direct 32bit link to the site. Followed that link we get a flag.
Mary Queen of Scots goes chinese. We capture secret message from prison where Mary Queen stands. Help us figure out what message means.
Guys, we were happy to get a Moomin's DNA! And we wonder, what substring is the origin of replication in the Moomin's DNA? Could you help us?
Universal dangerous positive: 126.96.36.199:1024. Send me your password: "3k8bbz032mrap75c8iz8tmi7f4ou00". Flag format is "RUCTF_.*"
In this task we are given a cpp source file with many libraries used and without any includes, you can find source here. So, the problem is pretty clear - find all the libraries, install and incldue them correctly.
In this task we have two similar messages of the form x + “Alex” (we'll call it x1) and x + “Jane” (we'll call it x2) where “+” is concatenation. These messages are encrypted with the RSA cipher using the same key (the module and the public exponent are given). The resulting messages are also provided:
61be5676e0f8311dce5d991e841d180c95b9fc15576f2ada0bc619cfb 991cddfc51c4dcc5ecd150d7176c835449b5ad085abec38898be02d27 49485b68378a8742544ebb8d6dc45b58fb9bac4950426e3383fa31a93 3718447decc5545a7105dcdd381e82db6acb72f4e335e244242a8e0fb bb940edde3b9e1c329880803931c
9d3c9fad495938176c7c4546e9ec0d4277344ac118dc21ba4205a3451 e1a7e36ad3f8c2a566b940275cb630c66d95b1f97614c3b55af860949 5fc7b2d732fb58a0efdf0756dc917d5eeefc7ca5b4806158ab87f4f44 7139d1daf4845e18c8c7120392817314fec0f0c1f248eb31af153107b d9823797153e35cb7044b99f26b0
The task is simple—find x.
Note: scroll down for the English translation.
Формат флага — «RUCTF_.*»
In this task we were given a fragment of the server-side code:
buf = c.recv(4096) digest, msg = buf.split(" ", 1) if (digest == md5(password+msg).hexdigest()): #here I send a secret else: c.send("Wrong signature\n")
It is also known that this string passes the test: “b34c39b9e83f0e965cf392831b3d71b8 do test connection” Unfortunately, this message is filtered and won't work, so we need to create a new correct message.
2 files are given:
Task name hints us to start with searching information about capturing GSM traffic.
Second link in google leads us to the great GSM write up written by Domi: http://domonkos.tomcsanyi.net/?p=418