anonymous
  • anonymous
Using RSA algorithm, what is the value of cipher text C, if the plain text M = 5 and p = 3, q = 11 & d = 7? A) 33 B) 5 C) 25 D) 26 Answer: (D) Please provide solution for this problem......
Computer Science
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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mathmate
  • mathmate
It is completely described in https://en.wikipedia.org/wiki/RSA_(cryptosystem) We are given: M=5 (plain text) p=3, q=11 (secret factors) d=7 (private key exponent) e=(1,\(\phi(n)\) ) such that e and \(\phi(n)\) are coprime AND d.e\(\equiv\)1 (mod n) public key is (n,e) To encrypt, \(c=M^e (mod~~n)\) where M is plain text, c=encrypted text To decrypt, \(M=c^d (mod~~n)\) We first calculate n=pq=33 \(\phi_n(n)=(p-1)(q-1)=20\) d=7, n=20, choose e=3 so that 7*3=21\(\equiv\)1 mod(20) check: 3 is between 1 and 20, and is coprime with 20. so Public key = (n,e) = (33,3) Private key = d = 7 Plain text = M = 5 encrypted text, c = M^e (mod n) = 5^3 mod (33) = 125 mod 33 = 26 decryption: M' = c^e mod n = 26^7 mod 33 = 5 = M (original plain text)
anonymous
  • anonymous
Thank you so much☺️
mathmate
  • mathmate
You're welcome! :)

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anonymous
  • anonymous
But I have a problem I don't know how they calculate mod in last can u please explain
mathmate
  • mathmate
Mod is the calculation of the remainder when one integer is divided by another. For example, 45 mod 11 equals 45-4*11=1 A more difficult example when we deal with big numbers: To find 26^7 mod 33 is to calculate the remainder when 26^7 is divided by 33. If you have a good calculator, you can find 26^7=8031810176. If your calculator does not show that many digits, just type in 26^7 in google, it will give you the answer. Then you divide 8031810176 by 33 to get 243388187.1515152 Take the integer portion 243388187 and multiply by 33 to get 8031810171. Subtract from 8031810176 to get 26^7 mod 33 = 5. You can also type in directly in Google 26^7 mod 33, it will return 5.
anonymous
  • anonymous
Thankyou so much ☺️☺️☺️
mathmate
  • mathmate
You're welcome! :)

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