634bdec633c5d4425e2cd8ee

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#!/usr/bin/env python3
# Link: https://crackmes.one/crackme/634bdec633c5d4425e2cd8ee
# The binary itself contains only a single small function, personally got fooled by
# the crackme's rating (3.3) as this is a really easy one. The algorithm does bitwise
# XOR with 0xa3 for certain characters to reveal the flag. As the whole binary is small,
# it's relatively easy to find a bunch of undefined bytes in the middle of it. Those
# are the bytes that can be used to brute force the flag.
data = [
0xBB,
0x96,
0x81,
0x96,
0xD3,
0x9A,
0x80,
0xD3,
0x8A,
0x9C,
0x86,
0x81,
0xD3,
0x95,
0x9F,
0x92,
0x94,
0xD3,
0xC9,
0xD3,
0xA1,
0xC1,
0xA5,
0xC1,
0xA1,
0xC6,
0xBA,
0xBD,
0xC6,
0xAC,
0xC7,
0xA0,
0xAC,
0xA1,
0xC7,
0xC1,
0xBF,
0xBF,
0xC4,
0xAC,
0xB5,
0xC1,
0xBD,
]
for k in range(256):
print(f"{k}, ", end="")
for d in data:
print(f"{chr(d ^ k)}", end="")
print("")
# Now it's simple to pick up the only sensible string from the output:
# k=243 (0xf3), Here is your flag : R2V2R5IN5_4S_R42LL7_F2N