zsombii's Easy crackme

Author:
zsombii

Language:
Java

Upload:
2025-12-13 15:43

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Platform:
Multiplatform

Difficulty:
1.3

Quality:
4.1

Arch:
java

Downloads:
1525

Size:
1.37 KB

Writeups:
6

Comments:
7

Description

My first crackme. Just find a valid validation key :)

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Comments (7)
Writeups (6)

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nightxyz on 2025-12-15 17:27: Too easy.

crazymoose on 2026-01-22 16:44: Straightforward but fun

idanhajbeko on 2026-02-17 20:28: There are 213 octillion valid keys. If you guess random 10 character keys, you should get a valid one after just 7 tries :|

AdrianS4957Y on 2026-02-20 20:52: That was easy

leizi999 on 2026-05-13 09:28: import java.util.Scanner; public class Main { public static boolean checkValidity(String s) { s.startsWith("KEY"); int v1 = 0; for(int v = 0; true; ++v) { int v2 = 1; if(v >= s.length()) { break; } if((s.charAt(v) & 3) != 0) { v2 = 0; } v1 += v2; } return v1 == 4; } public static void main(String[] arr_s) { Scanner scanner0 = new Scanner(System.in); System.out.print("Thanks for downloading this product. "); do { System.out.print("Enter the validation key here to complete, or exit with ENTER: "); String s = scanner0.nextLine(); if(s.trim().isEmpty()) { System.out.println("Exiting.."); System.exit(0); } if(s.length() != 10) { System.out.println("Length requirement not met"); } } while(s.length() != 10); if(Main.checkValidity(s)) { System.out.println("Validation key accepted"); return; } System.out.println("Validation key is invalid"); System.exit(0); } }

0xTh3B00Gym4n on 2026-07-21 03:32: [Click to reveal]

Ware0X on 2026-08-26 21:33: First time encountering bitwise ops used this way in Java! After a bit of research on how & 3 masks the ASCII values, I managed to reverse it and generated a working key: 8888KEYfeb (and realized the startsWith check was totally bypassed!). Super fun challenge (:

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Solution by 5l1m on 2025-12-16 14:53:

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Solution by SenorGPT on 2025-12-17 17:01:
Java `.jar` crackme. Using JADX, I recovered `checkValidity()` which scores the 10-character key by counting how many characters satisfy `(ch & 3) == 0` (i.e., ASCII divisible by 4). A key is valid if the score is exactly 4. I also wrote a small Python keygen to generate valid keys.

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Solution by 0xDuB on 2025-12-21 14:05:

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Solution by LogRob01 on 2026-01-17 22:18:
Java crackme solved using static analysis. The JAR file was decompiled with CFR to analyze the validation logic. No patching or keygen was used. The solution is based on understanding the character-based validation conditions.

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Solution by idanhajbeko on 2026-02-17 20:25:
explaining everything step by step uses --math-- to determine how many valid keys (spoiler: a lot!)

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Solution by Ware0X on 2026-08-27 13:29:
**Target:** Reverse-engineer `easycrack.jar` to get true keys * **Key Findings:** * **Length Constraint:** The length of key must be ten. * **Logic Bug:** `key.startsWith("KEY")` is

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zsombii's Easy crackme