634bdec633c5d4425e2cd8ee

thephilosopher-bruteverse/thephilosopher-bruteverse.py

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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