48 lines
1.1 KiB
Markdown
48 lines
1.1 KiB
Markdown
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## RSA Beginner
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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 `rsa (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: 3
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c: 219878849218803628752496734037301843801487889344508611639028
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n: 245841236512478852752909734912575581815967630033049838269083
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```
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#### Step-3:
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We run a simple `Flag.py` script to get the flag.
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```
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import gmpy2
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from gmpy2 import mpz
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e=mpz(3)
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c=mpz(219878849218803628752496734037301843801487889344508611639028)
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n=mpz(245841236512478852752909734912575581815967630033049838269083)
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#use factordb
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q=mpz(416064700201658306196320137931)
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p=mpz(590872612825179551336102196593)
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phi=gmpy2.mul(p-1,q-1)
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d=gmpy2.invert(e,phi)
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f=gmpy2.powmod(c,d,n)
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g=bytes.fromhex(hex(f)[2:])
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print("[+] Flag is : ",g)
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```
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#### Step-4:
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The output of `python3 Flag.py` is as follows:
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`[+] Flag is : b'abctf{rs4_is_aw3s0m3}'`
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#### Step-5:
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Finally the flag becomes:
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`abctf{rs4_is_aw3s0m3}`
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