65 lines
1.4 KiB
Markdown
65 lines
1.4 KiB
Markdown
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## RSA Noob
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The main idea finding the flag using RSA function and its operations.
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#### Step-1:
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After we download `rsanoob(1).txt` from the cloud, we try to understand what is the content.
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If anyone is unaware of RSA Encryption, they can checkout here:
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https://en.wikipedia.org/wiki/RSA_(cryptosystem)
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#### Step-2:
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The contents of `rsa (1).txt` are as follows:
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```
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e: 1
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c: 9327565722767258308650643213344542404592011161659991421
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n: 245841236512478852752909734912575581815967630033049838269083
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```
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#### Step-3:
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'e' and 'n' - Public key.
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'c' - Cipher text.
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Given `n` to us, we try to find `p` & `q` online from http://factordb.com/index.php
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#### Step-4:
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We get the `p` & `q` online as follows:
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<img src="Factorize.png">
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Now we can feed inputs manually to yield flag.
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#### Step-5:
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Running this `Flag.py` script:
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```
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from Crypto.Util.number import inverse
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import binascii
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e = 1
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c = 9327565722767258308650643213344542404592011161659991421
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n = 245841236512478852752909734912575581815967630033049838269083
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# From factordb
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p = 416064700201658306196320137931
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q = 590872612825179551336102196593
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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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hex_str = hex(m)[2:] # Removing '0x'
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print(binascii.unhexlify(hex_str))
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```
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#### Step-6:
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We get the flag by `python3 Flag.py`
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We get this output:
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`b'abctf{b3tter_up_y0ur_e}'`
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#### Step-7:
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Finally the flag becomes:
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`abctf{b3tter_up_y0ur_e}`
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