Cryptography - No need for brutus

Sat, 26 Apr 2025 17:51:33 +0530

No need for Brutus

Challenge Statement

Author:@aenygma

A simple message for you to decipher:

squiqhyiiycfbudeduutvehrhkjki

Submit the original plaintext hashed with MD5, wrapped between the usual flag format: flag{}

Solution

The title includes Brutus who is known for betraying Julius Caesar. Caesar happen to be named after a cipher he created, The Caesar cipher, also known as ROT cipher.

This suggests that the cipher provided might be a ROT cipher for which we don’t know the key for.

Using Cyberchef we can use the recipe ROT13 Brute Force to list all possible plaintext for the cipher.

We find a plain text with key value of 10. Since the challenge requires us to use the MD5Sum of this plaintext we can calculate it using:

echo -n "caesarissimplenoneedforbrutus" | md5sum

Covering the obtained hash with flag format yields the flag

Cryptography - Strive Marish Leadman TypeCDR

Sat, 26 Apr 2025 17:51:33 +0530

Strive Marish Leadman TypeCDR

Challenge Statement

Author:@aenygma

Looks like primo hex garbage.

Maybe something went wrong?

Can you make sense of it?

Note: This challenge was accompanied with a per-user instance

Solution

On spinning up the per-user instance, I was given a command to use Netcat into the instance.

On connecting it just print out the following data and exitted.

Now if you have worked with cryptosystems you could have recognized what this is right away. But those who don’t, this is variable for a RSA public cryptosystem.

The given are everything that needed to encrypt and decrypt data.

A quick run down of that is:

p and q are supposed to be very large prime numbers
n = p * q
Now a value called Euler’s phi function for the value n is calculated as phi(n) = (p-1)*(q-1)
Now a random value e is selected such that e < phi(n) and e and phi(n) are co-prime
This e is the public key
Now the corresponding private key is derived such that d*e = 1 mod phi(n)
With this one can encrypt message m to cipher c with public key as c = (m^e) mod n
This can be reversed or decrypted using private key as m = (c^d) mod n

Since we have all the variables and assuming the data in the bottom is ciphertext, this boils down to a simple decrytion.

I used a tool called rsactftool to get it done.

By just plugging in the values of p, q, d, e and data to decrypt, the tool even decoded the output to UTF-8 making my job easier.

Cryptography on Monish Kumar's Blog

Cryptography - No need for brutus

No need for Brutus

Challenge Statement

Solution

Cryptography - Strive Marish Leadman TypeCDR

Strive Marish Leadman TypeCDR

Challenge Statement

Solution