76 lines
1.9 KiB
Markdown
76 lines
1.9 KiB
Markdown
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# RAID Safety Assays, But Fixed
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> The flag to this challenge is all lowercase, with no underscores.
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> e = 65537
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> n = 4629059450272139917534568159172903078573041591191268130667
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> c = 6743459147531103219359362407406880068975344190794689965016
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## About the Challenge
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We were given a python code and here is the content of the file
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```python
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from Crypto.Util.number import *
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import random
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p = getPrime(96)
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q = getPrime(96)
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n = p*q
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e = 65537
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flag = b'ctf{0000000000000000}'
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flag = str(pow(bytes_to_long(flag), e, n))
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perm = list(range(10))
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random.shuffle(perm)
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perm = list(map(str, perm))
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c = ''.join([perm[int(x)] for x in flag])
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print(f'e = {e}')
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print(f'n = {n}')
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print(f'c = {c}')
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```
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The aboce code generates two prime numbers `p` and `q`, calculates the modulus `n` for RSA encryption, encrypts a flag using the RSA encryption algorithm with the public exponent `e`, and applies a permutation cipher to the encrypted flag. The resulting encrypted flag `c` is then printed along with the values of `e` and `n`.
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## How to Solve?
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I created a python code to performs a brute-force search over all possible permutations of digits from 0 to 9 and I got the value of `p` and `q` using FactorDB
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```python
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import itertools
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c = '6743459147531103219359362407406880068975344190794689965016'
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e = 65537
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p = 62682123970325402653307817299
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q = 73849754237166590568543300233
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permutations = list(itertools.permutations(range(10)))
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for perm in permutations:
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perm = list(map(str, perm))
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c_list = [int(x) for x in c]
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inv_perm = [perm.index(str(i)) for i in range(10)]
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plaintext_permuted = [inv_perm[x] for x in c_list]
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c_rev = ''.join(map(str, plaintext_permuted))
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n = p*q
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phi = (p-1)*(q-1)
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d = inverse(e,phi)
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m = pow(int(c_rev),d,n)
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print(long_to_bytes(m))
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```
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And then run this program like this
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```shell
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python3 solve.py | grep "ctf"
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```
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```
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ctf{cryptpainfulflag}
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```
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