RSA Beginner
50 points Medium

I found this scribbled on a piece of paper. Can you make sense of it? https://mega.nz/#!zD4wDYiC!iLB3pMJElgWZy6Bv97FF8SJz1KEk9lWsgBSw62mtxQg

Flag
Cryptography ยท intelagent
3898 solves
Rating 4.49
5
4
3
2
1
Must solve to rate

Discussion

avatar
Dash_Line

2 months ago

Reply Share

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.

0
avatar
HLx00

5 months ago

Reply Share

any thing ez with py

1
avatar
woods

5 months ago

Reply Share

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

1
avatar
Selrahc

6 months ago

Reply Share

the flag is ____

0
avatar
el_zoz_22

10 months ago

Reply Share

.

0
avatar
el_zoz_22

10 months ago

Reply Share

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))

-2
avatar
FISHBURRITO

11 months ago

Reply Share

this should be in easy

2
avatar
bloman

1 year ago

Reply Share

A good challenge that help us to write a python script

0
avatar
n00bz

1 year ago

Reply Share
Protected
-4
avatar
dimuthsakya

1 year ago

Reply Share

Easy peazy

0
avatar
H4cked

3 years ago

Reply Share

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

2
avatar
one1

3 years ago

Reply Share

You have to convert this value into ASCII text

0