Tags: rev apk

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# THREE LITTLE KEYS

This challenge gives us an apk file. After opening it, it showed us 3 keys. The first one instantly turns green after clicking it, the second one wants us to enter something and the third one throws an error.

## First Key

I decompiled the .dex file and got many folders to search through. I opened the most important main class for the app in de/cybersecurityrumble/threelittlekeys and took a closer look at how the app actually works.
After clicking a button, the corresponding method in another class gets run. I found that class in c/a/a/c and quickly saw the method for the first key.

Java
public boolean checkFirstKey() {
String num = Integer.toString(new Random(3762).nextInt(8000));
this.f1665a = num;
Log.i("Key Created", num);
return !this.f1665a.isEmpty();
}


As we can see, the method simply generates a random number and then prints it to the console... It seems to use the same seed every time so I tried to run that piece of code myself and got the first key which is 7765. I confirmed that by taking a look at the log while running the app in the android studio emulator.

## Second Key

The next step was to find the second key. Somehow I was not able to decompile this method using the usual decompiler tool websites. I then tried jadx-gui which somehow managed to decompile the method as good as possible, I guess it was corrupted on purpose.

The method gave us a few clues to build our second key:

The second character is a "c", the 1., 3., 6. and 7. character are the same and there is another method that converts the string into a byte array, after testing I figured out that it converts a string with hex values to the corresponding byte values.

That byte array also had the requirement, that the last character is an e and that it is 4 characters long. If we reverse that, we can find out that the key is 8 characters long and that it ends with 65, as the hex value for e is 65.
The last thing needed to make that key valid is that its byte array characters have equal 100 when bitwise and'ed together. Doing this I got around 20 possible combinations and after decoding it to ascii I got love in one of the results which had to be the key, or at least the hex version which is 6c6f7665

## Third Key

This time I had no problems decompiling the method. It uses a regex and a few comparisons to validate the key. The regex \d{2}[*!=()%?$#]\p{Upper}\d+! matches a string that starts with two digits, followed by any of the characters *!=()%?$#, followed by an uppercase letter, followed by one or more digits, and finally an exclamation mark. Also, the 1. and 7. character have to be equal and the 2. and 5. have to do the same.

But how am I going to give the key to the app?

I had to take a closer look at how the app gets the flag in the first place.

In the check method it says:
java
String string = d.a().f1669a.getString("ThirdKey", "");


I went to class d and looked at the right method and I now know how the app gets the key. It is stored in the SharedPreferences!

But how do I get access on it?

I extracted the app using apktool, then I added android:debuggable="true" to the AndroidManifest.xml and after that I built the app again and also had to sign it to make it installable. Now I am able to not only read the file in the file explorer of android studio, but also to edit it using the adb shell.

I correctly put a guess 12*A221! into the file and was able to successfully make the last key green. Still, something has to be wrong as clicking the unlock button throws an error... what if I actually had to find the correct third key?

After putting the openLock() method into a java project I wrote a small bruteforcer that tries all of the combinations.

java
public static void openLock() {
String str = "7765";
String str2 = "6c6f7665";
String str3 = "12*A221!"; // Bruteforce
String string = "2a";
for (int i = 0; i < 10; i++) {
for (int ii = 0; ii < 10; ii++) {
for (String e : "*!=()%?\$#".split("")) {
for (String f : "ABCDEFGHIJKLMNOPQRSTUVWXYZ".split("")) {
for (int iii = 0; iii < 10; iii++) {
str3 = "" + i + "" + ii + e + f + "" + ii + "" + iii + "" + i+"!";
byte[] a2 = a(str + string + str2 + string);
byte[] bytes = str3.getBytes();
byte[] bArr = new byte[(a2.length + bytes.length)];
System.arraycopy(a2, 0, bArr, 0, a2.length);
System.arraycopy(bytes, 0, bArr, a2.length, bytes.length);
SecretKeySpec secretKeySpec = new SecretKeySpec(bArr, "AES");
try {
byte[] bArr2 = FLAG;
instance.init(2, secretKeySpec);
if(new String(instance.doFinal(bArr2), StandardCharsets.UTF_8).startsWith("CSR")){
System.out.println(new String(instance.doFinal(bArr2), StandardCharsets.UTF_8) + " with keys: \n1. "+str+"\n2. "+str2+"\n3. "+str3);
}
} catch (Exception ex) {
}
}
}
}
}
}
}


The encrypted FLAG was at the top of that class and I got the variable "string" from the config.

And one of the combinations gave me the correct flag!

Original writeup (https://github.com/xXLeoXxOne/writeups/blob/main/CyberSecurityRumble%202022/THREE%20LITTLE%20KEYS.md).