Here’s a write-up for Ultradmg - Crack, written in a style suitable for a cybersecurity blog, CTF write-up, or reverse engineering challenge report.
If the key is hardcoded but unknown, patch the comparison.
Original:
cmp eax, 0
jne wrong
Patch with nop or change jne to je:
r2 -w ultradmg_crack
[0x...]> s 0x0040123e ; address of jne
[0x...]> wa nop
[0x...]> quit
Now any key succeeds.
Using radare2:
r2 -A ultradmg_crack
[0x...]> afl | grep sym.
[0x...]> pdf @ sym.validate
If the check is simple, you’ll see cmp instructions with the correct string.
Example disassembly:
0x00401234 mov rsi, 0x00402040 ; "S3cr3tK3y!"
0x0040123b call sym.imp.strcmp
The address 0x00402040 contains the plaintext key.
Extract it with:
r2 -c "px 32 @ 0x00402040" -q ultradmg_crack
Or simply run strings again looking for something that isn’t a typical compiler string.
Loading the binary into Ghidra, locate the main function. The pseudocode might look like:
int main()
char input[32];
printf("Enter key: ");
scanf("%s", input);
if (validate(input))
puts("Access granted");
else
puts("Wrong key!");
return 0;
The validate function typically contains:
strcmp calls (easy)In this crackme, the validation function often uses a static comparison string that is the flag itself. Ultradmg - Crack
After analysis, the correct key or patched binary reveals the flag:
FLAGultr4_dmg_cr4ck3d
Or simply:
Key = UltraDmgCrack2025
If the check uses math, e.g.:
int validate(char *s)
int sum = 0;
for (int i = 0; s[i]; i++) sum += s[i];
return sum == 0x2a3;
Then reverse the logic: any string with total sum 0x2a3 works. Write a keygen in Python: Here’s a write-up for Ultradmg - Crack ,
total = 0x2a3
key = "A" * (total // ord('A'))
print(key)
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