HITCON CTF 2015 Quals rsabin writeup
HITCON: participated as a member of fuzzi3, solved just one chall: rsabin (crypto 314)
— Ikumi Shimizu (@_193s) 2015, 10月 18
rsabin (Crypto 314)
author: 193s (fuzzi3)
Python script rsabin.py
and an encrypted flag was given.
The modulo, N was 314bit, so first of all I started factoring N using yafu.
It took over 70 minutes…
p = 164184701914508585475304431352949988726937945291
q = 123722643358410276082662590855480232574295213977
Looks good, there we go!
rsabin = rsa + rabin
It seems like you just need to calculate m = c^d (mod N)
where private key, d = mod_inv(e, φ(N))
, but it won’t work: In this case, mod_inv(e, φ(N))
can’t be calculated! Yes, e, RSA’s public exponent must be chosen as coprime to φ(N).
e = 31415926535897932384 = 2^5 * 563 * 15647 * 93967 * 1186001
φ(N) = (p1)(q1) = 2^4 * 3^2 * 5 * 13 * 3769 * 27077 * 17846641 * 83901947 * 159936353 * 310427287 * 95195293927205442569 * 3004887825587533352438332710071
The challenge title rsabin
reminded me of rsa
and rabin
!
The encryption algorithm for rabin cipher (in case k=0) is as follows:c = m^2 mod N
I was certain that I can calculate m^(e/2) mod N
from m^e mod N
when e is an even number.
So, what I need to do is:
 apply this method 5 times, then I get
c' = m^e' mod N
where e’ = e/32 (satisfiesgcd(e', φ(N)) = 1
)  decrypt
m
with normal RSA:d' = mod_inv(e', φ(N))
,m = c'^d' mod N
.
Note that decryption algorithm for rabin cipher returns 4 numbers (m1, m2, m3, m4), and the right plaintext is one of them (there is no way to tell which is the correct one). Because of this, there are 4**5=1024 possible ways of c'
.
bruteforcing missing info
I calculated m
by using above method for 1024 possible combinations (actually only 16 unique values satisfied m'^e = c
), but none of them looked like a flag.
rsabin.py:


What?! FLAG is 400bits length but N is 314bits?
ok, that means that some upper bits of the flag are missing. (now I have m = FLAG mod N
, so just need to know i
satisfies FLAG = iN+m
)
( I contacted the admin and made sure that the flag is not in the following format: \x00\x00\x00...\x00hitcon{....}
)
We know that the flag starts with hitcon{
, followed by asciiprintable characters, and ends with }
, so I tried to bruteforce i
(satisfies FLAG = iN+m
) to find the flag matches ^hitcon{[:print:]+}$
for each m.
I wrote a little script, and after 20 minutes, my script showed me the flag!
flag: hitcon{Congratz~~! Let's eat an apple pi <3.14159}