## CR3 What is this encryption?

Crypto - 150 Points

Fady assumed this time that you will be so n00b to tell what encryption he is using he send the following note to his friend in plain sight :

``````p=0xa6055ec186de51800ddd6fcbf0192384ff42d707a55f57af4fcfb0d1dc7bd97055e8275cd4b78ec63c5d592f567c66393a061324aa2e6a8d8fc2a910cbee1ed9
q=0xfa0f9463ea0a93b929c099320d31c277e0b0dbc65b189ed76124f5a1218f5d91fd0102a4c8de11f28be5e4d0ae91ab319f4537e97ed74bc663e972a4a9119307
e=0x6d1fdab4ce3217b3fc32c9ed480a31d067fd57d93a9ab52b472dc393ab7852fbcb11abbebfd6aaae8032db1316dc22d3f7c3d631e24df13ef23d3b381a1c3e04abcc745d402ee3a031ac2718fae63b240837b4f657f29ca4702da9af22a3a019d68904a969ddb01bcf941df70af042f4fae5cbeb9c2151b324f387e525094c41
``````

He is underestimating our crypto skills!

### Writeup

`p`, `q`, `e`, `c`. Just casual name if you don’t know the most studied public-key cryptosystem RSA.

The security of RSA holds as long as the attacker can’t factorize `n` (that is equals to `p*q`).
Here Fady commited a big mistake by giving us `p` and `q`. With those we can compute `λ(n)` using the Carmichael’s totient function with `lcm(p-1,q-1)` and finally find the decryption key `d`, the modular multiplicative inverse of e (modulo λ(n)) -> `d⋅e ≡ 1 (mod λ(n))`

So, this can be easily done with p4’s crypto-commons library

Magic

The flag is `ALEXCTF{RS4_I5_E55ENT1AL_T0_D0_BY_H4ND}`