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Mini RSA

What happens if you have a small exponent? There is a twist though, we padded the plaintext so that (M ** e) is just barely larger than N. Let's decrypt this: ciphertext

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 we can actually keep on adding multiples of onto 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.

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}

Mini RSA

What happens if you have a small exponent? There is a twist though, we padded the plaintext so that (M ** e) is just barely larger than N. Let's decrypt this: ciphertext

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}