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
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
this should be in easy
8 months ago
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.