48 lines
1018 B
Markdown
48 lines
1018 B
Markdown
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## little RSA
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The main idea finding the flag is getting the cipher text from RSA algorithm.
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#### Step-1:
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After I downloaded `a.txt` & `flag.zip`, I checked out the contents in them.
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`a.txt` gave `c`, `n`, `e` as follows:
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```
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c=32949
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n=64741
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e=42667
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```
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`flag.zip` contains `flag.txt` which is encrypted by a pin which is key from RSA implementation.
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#### Step-2:
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So, I used again the [RsaCtf Tool](https://github.com/Ganapati/RsaCtfTool) and implemented by a `flag.py`:
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`n` was factorized online at http://factordb.com/index.php?query=64741 to get `p` & `q`.
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```python
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from Crypto.Util.number import inverse
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import binascii
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e = 42667
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c = 32949
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n = 64741
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# From factordb
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p = 101
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q = 641
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phi = (p-1) * (q-1)
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d = inverse(e,phi)
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m = pow(c,d,n)
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print (m)
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```
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#### Step-3:
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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`.
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Voila! I got the flag!
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#### Step-4:
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Finally the flag becomes:
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`csictf{gr34t_m1nds_th1nk_4l1ke}`
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