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Challenge "Broadcasting and low exponent — RSA Attack"  

  By: admin on July 22, 2011, 8:03 p.m.

A message was encrypted with three different moduli but the same public exponent. Fortunately, you were able to catch the three different ciphertexts. Are you able to recover the secret message?
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 Last edited by: admin on July 22, 2022, 6:11 p.m., edited 2 times in total.

Re: Challenge "Broadcasting and low exponent - RSA-Attack"  

  By: nakul22 on Feb. 12, 2012, 1:29 a.m.

Hi
I am interested to do this challenge but i didnt get the exact idea of this challenge so please give me brief idea of this topic and at least tell me how to proceed.

Regards
Govind

Re: Challenge "Broadcasting and low exponent - RSA-Attack"  

  By: aurelie on Feb. 24, 2012, 8:35 p.m.

I think page 5 is a pretty good hint. Maybe you can do some investigations on low exponents for RSA and the use of the CRT.

Re: Challenge  

  By: kiekuk on March 31, 2015, 6:02 p.m.

Hello,

i have a problem with this challenge. I used the chinese remainder theorem, but the solution is maybe wrong.

Can someone help me?

Thank you!

Best regards
kiekuk

Re: Challenge  

  By: be on April 1, 2015, 1:07 p.m.

i have a problem with this challenge. I used the chinese remainder theorem, but the solution is maybe wrong.

Hello kiekuk, we noticed, you solved the challenge in the meantime. Could you please add a post, what information or idea was helpful or where your fault was. This might be interesting for other solvers, too. Best regards, be

Decrypted Decimal to plaintext ?  

  By: pratikvpatil on Oct. 5, 2017, 10:12 a.m.

I have cracked the challenge and got a decimal number for plaintext. I can be sure of this because I encrypted my message with the exponent and the moduli to get the ciphertext which is given in the challenge. However, I cannot understand how to interpret the number back to plaintext to verify my answer on this website. I would be really grateful if someone can help me here.

Re: Decrypted Decimal to plaintext ?  

  By: be on Nov. 5, 2017, 1:55 p.m.

I have cracked the challenge and got a decimal number for plaintext … However, I cannot understand how to interpret the number back to plaintext

Hello pratikvpatil.
Thanks for your question. You are almost done.
Try to represent the final number in hex and see if you can make sense of it. For example, a hex value of 47 means the letter "G".

Re: Challenge  

  By: Majakovskij on Oct. 3, 2020, 12:12 p.m.

Hi all. It seems to me that I've found the plaintext (in German language), but as I try to submit the entire text (just copying it from plaintext) it get rejected. I've also tried to add some punctuation mark, but obviously it didn't make the trick. What am I missing?

Re: Challenge  

  By: Majakovskij on Oct. 4, 2020, 10:23 a.m.

Hi all. It seems to me that I've found the plaintext (in German language), but as I try to submit the entire text (just copying it from plaintext) it get rejected. I've also tried to add some punctuation mark, but obviously it didn't make the trick. What am I missing?

Problem solved. The problem was the actual representation of the integer value of the message . After "the computation" it actually appeared to be a decimal value (it may probably depend on the library you're using, I didn't figure it out yet). So if you get an intelligible plaintext you may wanna try to round up the integer value (since it should only change the last bit). It did the work for me and I hope it will do it for you as well.


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