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

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## 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](https://github.com/Ganapati/RsaCtfTool) and implemented by a `flag.py`:
`n` was factorized online at http://factordb.com/index.php?query=64741 to get `p` & `q`.
```python
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}`