June 19, 20187 yr Difficulty : 5 Language : .NET Platform : Windows OS Version : All Packer / Protector : ConfuserEx - Custom Description : This is my first time as a newbe; I hope I didnt do too badly 😅. Try to find at least one valid alphanumeric Key. The key has the following format: XXXX-XXXX-XXXX-XXXX-XXXX-XXXX, it can have letters and numers. Good luck! Screenshot : KeygenMe1.zip
June 20, 20187 yr 1/10, almost nothing moded Spoiler private void button1_Click(object sender, EventArgs e) { this.textBox1.Text.ToCharArray(); if (this.textBox1.Text.Length == 29) { if (this.textBox1.Text.Contains("-") && this.textBox1.Text.ToCharArray()[4] == '-' && this.textBox1.Text.ToCharArray()[9] == '-' && this.textBox1.Text.ToCharArray()[14] == '-') { if (this.textBox1.Text.ToCharArray()[19] == '-') { if (this.textBox1.Text.ToCharArray()[24] == '-') { if (!this.method_0()) { MessageBox.Show("Sorry Duke, wrong key. Try again!"); return; } this.label2.Visible = true; return; } } } } MessageBox.Show("The entered Key doesnt have the correct format", ":(", MessageBoxButtons.OK, MessageBoxIcon.Exclamation); } KeygenMe1-cleaned.zip Edited June 20, 20187 yr by Reza-HNA
June 20, 20187 yr 2 hours ago, Reza-HNA said: 1/10, almost nothing moded Hide contents private void button1_Click(object sender, EventArgs e) { this.textBox1.Text.ToCharArray(); if (this.textBox1.Text.Length == 29) { if (this.textBox1.Text.Contains("-") && this.textBox1.Text.ToCharArray()[4] == '-' && this.textBox1.Text.ToCharArray()[9] == '-' && this.textBox1.Text.ToCharArray()[14] == '-') { if (this.textBox1.Text.ToCharArray()[19] == '-') { if (this.textBox1.Text.ToCharArray()[24] == '-') { if (!this.method_0()) { MessageBox.Show("Sorry Duke, wrong key. Try again!"); return; } this.label2.Visible = true; return; } } } } MessageBox.Show("The entered Key doesnt have the correct format", ":(", MessageBoxButtons.OK, MessageBoxIcon.Exclamation); } KeygenMe1-cleaned.zip Unpack is very easy you're right but how do keygen? I can't do that is that possible?
June 20, 20187 yr Author 13 hours ago, Reza-HNA said: 1/10, almost nothing moded Reveal hidden contents private void button1_Click(object sender, EventArgs e) { this.textBox1.Text.ToCharArray(); if (this.textBox1.Text.Length == 29) { if (this.textBox1.Text.Contains("-") && this.textBox1.Text.ToCharArray()[4] == '-' && this.textBox1.Text.ToCharArray()[9] == '-' && this.textBox1.Text.ToCharArray()[14] == '-') { if (this.textBox1.Text.ToCharArray()[19] == '-') { if (this.textBox1.Text.ToCharArray()[24] == '-') { if (!this.method_0()) { MessageBox.Show("Sorry Duke, wrong key. Try again!"); return; } this.label2.Visible = true; return; } } } } MessageBox.Show("The entered Key doesnt have the correct format", ":(", MessageBoxButtons.OK, MessageBoxIcon.Exclamation); } KeygenMe1-cleaned.zip Nice, I knew youd easily unpack it, however this wasnt the entire purpose of this challenge. I'm asking if you can figure out at least one Key or better even write a keygen to this little app 😉
June 20, 20187 yr 1 hour ago, SaggingCoder said: I'm asking if you can figure out at least one Key or better even write a keygen to this little app Given the way you designed this app, I believe it is not keygenable. Basically, what you're asking is this: "knowing the encrypted data and the original data, find the used encryption key". It's called known-plaintext attack and it's practically impossible on modern algorithms like AES (when they are used in the way you did). However, I'll be happy to be proven wrong..
June 20, 20187 yr Author 1 hour ago, kao said: Given the way you designed this app, I believe it is not keygenable. Basically, what you're asking is this: "knowing the encrypted data and the original data, find the used encryption key". It's called known-plaintext attack and it's practically impossible on modern algorithms like AES (when they are used in the way you did). However, I'll be happy to be proven wrong.. Oooooh ok, I understand and I apologize! I thought there still might be a way to somehow reverse the rijndael algorithm... (or if it's impossible, just bruteforcing until one key is found) Welp then here is one key: 8C08-UT01-70J9-9ST3-D416-09A0, writing a keygen shouldnt be too hard now Edited June 20, 20187 yr by SaggingCoder
June 21, 20187 yr Solution ..and with one known key it's perfectly solvable. All credits go to @Reza-HNA for deobfuscating the keygenme. After that, it was a piece of cake. Keygen is not obfuscated in any way, so anyone can take a look how it's done. keygen for Newbe KeygenMe1.zip
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