# Mini RSA

Now we're getting onto proper cryptography. Here we are told that we are using RSA with a small exponent, but $$M^e$$ is just more than $$N$$. This means we can't **quite** do a [cube-root attack, but because it is **just** more than $$M$$ we can actually keep on adding multiples of $$N$$ onto $$c$$ ](/notes/cryptography/overview/public-exponent-attacks/small-e.md)and taking the cube root until we get the flag. As it is close, it's not infeasible to brute force. I use gmpy2's `iroot` function to take the cube root.

```python
from Crypto.Util.number import long_to_bytes
from gmpy2 import iroot

N = 1615765684321463054078226051959887884233678317734892901740763321135213636796075462401950274602405095138589898087428337758445013281488966866073355710771864671726991918706558071231266976427184673800225254531695928541272546385146495736420261815693810544589811104967829354461491178200126099661909654163542661541699404839644035177445092988952614918424317082380174383819025585076206641993479326576180793544321194357018916215113009742654408597083724508169216182008449693917227497813165444372201517541788989925461711067825681947947471001390843774746442699739386923285801022685451221261010798837646928092277556198145662924691803032880040492762442561497760689933601781401617086600593482127465655390841361154025890679757514060456103104199255917164678161972735858939464790960448345988941481499050248673128656508055285037090026439683847266536283160142071643015434813473463469733112182328678706702116054036618277506997666534567846763938692335069955755244438415377933440029498378955355877502743215305768814857864433151287
e = 3

c = 1220012318588871886132524757898884422174534558055593713309088304910273991073554732659977133980685370899257850121970812405700793710546674062154237544840177616746805668666317481140872605653768484867292138139949076102907399831998827567645230986345455915692863094364797526497302082734955903755050638155202890599808147130204332030239454609548193370732857240300019596815816006860639254992255194738107991811397196500685989396810773222940007523267032630601449381770324467476670441511297695830038371195786166055669921467988355155696963689199852044947912413082022187178952733134865103084455914904057821890898745653261258346107276390058792338949223415878232277034434046142510780902482500716765933896331360282637705554071922268580430157241598567522324772752885039646885713317810775113741411461898837845999905524246804112266440620557624165618470709586812253893125417659761396612984740891016230905299327084673080946823376058367658665796414168107502482827882764000030048859751949099453053128663379477059252309685864790106


for i in range(10_000):
    c_try = c + i * N
    m = int(iroot(c_try, 3)[0])
    flag = long_to_bytes(m)

    if b'pico' in flag:
        print(flag)

# picoCTF{e_sh0u1d_b3_lArg3r_7adb35b1}
```


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