CSICTF-Writeups/Crypto/little RSA/README.md

little RSA

The main idea finding the flag is getting the cipher text from RSA algorithm.

Step-1:

After I downloaded a.txt & flag.zip, I checked out the contents in them.

a.txt gave c, n, e as follows:

c=32949
n=64741
e=42667

flag.zip contains flag.txt which is encrypted by a pin which is key from RSA implementation.

Step-2:

So, I used again the RsaCtf Tool and implemented by a flag.py:

n was factorized online at http://factordb.com/index.php?query=64741 to get p & q.

from Crypto.Util.number import inverse
import binascii

e = 42667
c = 32949
n = 64741

# From factordb

p = 101
q = 641

phi = (p-1) * (q-1)

d = inverse(e,phi)
m = pow(c,d,n)

print (m)

Step-3:

After running above script as python3 flag.py, I got this output as 18429. I used this key to unlock the zip to get access to flag.txt.

Voila! I got the flag!

Step-4:

Finally the flag becomes: csictf{gr34t_m1nds_th1nk_4l1ke}