eventhorizon02's super

Author:
eventhorizon02

Language:
C/C++

Upload:
2021-08-19 17:59

Download

Platform:
Unix/linux etc.

Difficulty:
1.8

Quality:
4.8

Arch:
x86-64

Downloads:
6

Writeups:
7

Comments:
7

Description

My first upload, should be fairly easy to solve

Comments (7)
Writeups (7)

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NecroTechno on 2021-09-06 13:54: Definitely an interesting one! There are tons of solutions to this one.

aziz00 on 2021-09-11 20:17: thanks a lot , by the way try these password : lol Correct : 881 Correct : 809 Correct : 659 Correct : 761 Correct : 433 Correct : 659 Correct : 241511 Correct : 160141 Correct : 107273 Correct : 353076209 Correct : 276330823

Anhkhoa on 2021-09-19 03:03: very easy for beginer

blackfoot50115 on 2021-10-19 02:22: I submitted a solution. It has not yet been approved. Would you be able to tell me when it could be approved. If it was not approved would you be able to tell me why or what else I need to submit?

sumeet on 2022-06-18 12:01: 999853 999863 999883 999907 999917 999931 999953 999959 999961 999979 999983 and many more

0xAron on 2022-12-22 14:55: Done!

Nader_ayman on 2024-09-23 12:42: Spoiler ! Is it just a prime number that it's length is divisible by 3 ?! ____________________ congratulations you have cracked the code flag{103} flag{100003} flag{100000007}

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Solution by Kryptoenix on 2021-09-05 21:26:
Great chall!

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Solution by MANA624 on 2021-10-15 02:59:
Cool challenge! The underlying assembly is a little more complex than the corresponding logic, so be sure to use a decompiler!

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Solution by blackfoot50115 on 2021-10-16 01:25:

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Solution by ThePhilosopher on 2021-10-16 16:58:
a comprehensive solution with a little bit of details!

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Solution by ne0n_c4lic0 on 2022-04-20 15:21:
Walkthrough - Reversing and Python3 Auto Cracking

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Solution by 0xAron on 2022-12-22 14:54:
Ez soluton and automate using PYTHON3

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Solution by amnam on 2023-09-29 19:21:
Decompile using Ghidra and a sample flag.

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crackmes.one

crackmes.one

eventhorizon02's super_easy