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

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Cryptography ยท intelagent
4308 solves
Rating 4.50
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Discussion

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Dash_Line

8 months ago

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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.

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HLx00

11 months ago

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any thing ez with py

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woods

1 year ago

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

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Selrahc

1 year ago

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the flag is ____

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el_zoz_22

1 year ago

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.

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el_zoz_22

1 year ago

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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))

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FISHBURRITO

1 year ago

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this should be in easy

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bloman

2 years ago

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A good challenge that help us to write a python script

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0xl4p

2 years ago

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Protected
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dimuthsakya

2 years ago

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Easy peazy

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H4cked

4 years ago

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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

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one1

4 years ago

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You have to convert this value into ASCII text

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