ctf-writeup/CTF ARA 2023/Secrets Behind a Letter

Secrets Behind a Letter

Melon and Edith went to an labyrinth and they should break the code written on a letter in a box in order to escape the labyrinth.

Open the letter and break the code

About the Challenge

Given a file containing p, q, e, and c

p:
12575333694121267690521971855691638144136810331188248236770880338905811883
48506410486564983492781972561769555447210034136189616202231165330153281010
1344273
q:
12497483426175072465852167936960526232284891876787981080671162783561411521
67580911220457361735838974273254629350270958512920588572607849241710986751
2398747
c:
36062934495731792908639535062833180651022813589535592851802572264328299027
40641392734685245421762779331514489294202688698082362224015740571749978795
99430405407341221428388984827675412726778370913038246699129635727146561394
22011853028133556111405072526509839846701570133437746102727644982344712571
844332280218
e = 65537

The variable c is the encrypted flag. The variable e is the exponent, and p and q are the two prime numbers used to generate the private and public keys.

How to Solve?

To solve this problem, I created a python script like the image below (You can access the solver [here])

The variable n is the RSA modulus, which is the result of multiplying p and q. Next, the variable phi is calculated using Euler's totient function, namely (p-1)*(q-1). The variable d is then calculated using the inverse modulo of e and phi, which are the RSA private keys. And in the last line, the pow() function is used to calculate flags that have been encrypted using the public key, using the private key. The result is converted to bytes using the long_to_bytes() function and we get the flag.

ARA2023{1t_turn5_0ut_to_b3_an_rsa}