fun fact: RSA beginner is easier than RSA noob. hint (since you're in the comments,): use a famous french d-coding website for this one.

any thing ez with py

Hint: https://www.dcode.fr/rsa-cipher

the flag is ____

.

we can use {https://www.alpertron.com.ar/ECM.HTM} to get a,b from n

a=416064700201658306196320137931 __ b=590872612825179551336102196593

n=245841236512478852752909734912575581815967630033049838269083 ct=219878849218803628752496734037301843801487889344508611639028

puy=(a-1)*(b-1)

e=3

d=inverse(e,puy)

flag=pow(ct,d,n)

print(long_to_bytes(flag))

this should be in easy

A good challenge that help us to write a python script

Easy peazy

Why isn't this the flag?

P = 590872612825179551336102196593 Q = 416064700201658306196320137931 D = E^-1 mod ((P-1)(Q-1)) = 163894157674985901835273156607712429668627194783678277289707 M = C^D mod N = 219878849218803628752496734037301843801487889344508611639028^163894157674985901835273156607712429668627194783678277289707 mod(245841236512478852752909734912575581815967630033049838269083) = 142327357830311032682897538242625561166817486779261

You have to convert this value into ASCII text

this should be in easy

any thing ez with py

Hint: https://www.dcode.fr/rsa-cipher

RSA RSA RSA RSA ... RSA!

can anyone explain me how to convert from plain text to text pls i have watched some solution and i still dont understand which form of this string