Discussion
Cool! use python.
Cool! use python.
very cool little challenge. if you see the comments you'll see a bunch of people with differnet solutions they came up with. if you dont know how to program stuff and is lazy so you don't wanna download python3 to solve this, its easy. e=1 means that the private key, d, is also 1, if that happens, turn to the equation c = m^e mod n. e=1, so c = m mod n. ignore mod n for now, you'll see, it says that the OTHER private key, m, is actually equal to c. so to get the flag, you'll need to read the descriptions of RSA solvers and learn how it encrypts and decrypts stuff to and from RSA. you should only be using two translators and no more than that. (RSA solvers wont help so good luck)
Almost had it!
very cool little challenge. if you see the comments you'll see a bunch of people with differnet solutions they came up with. if you dont know how to program stuff and is lazy so you don't wanna download python3 to solve this, its easy. e=1 means that the private key, d, is also 1, if that happens, turn to the equation c = m^e mod n. e=1, so c = m mod n. ignore mod n for now, you'll see, it says that the OTHER private key, m, is actually equal to c. so to get the flag, you'll need to read the descriptions of RSA solvers and learn how it encrypts and decrypts stuff to and from RSA. you should only be using two translators and no more than that. (RSA solvers wont help so good luck)
c = m^e mod n
Here we have c, e and n.. we need to calculate m. Since e = 1 its very easy to solve. Final value has to be converted from declimal to hexa to text
Nice enough to do baby steps, but this challenge should come before RSA Beginner I think
nice
u should up ur e and n also :/
6 months ago
easiest ctf in the world