UGC NET CSE | December 2012 | Part 3 | Question: 8

Question

Using RSA algorithm, what is the value of cipher text C, if the plain text M=5 and p=3, q=11 and d=7?

• A. 33
• B. 5
• C. 25
• D. 26

Answer 1

Given p=3,q=11. n=(p X q) = (3 X 11)=33
   m=(p-1)X(q-1) = (2 X 10)=20
Find a small odd integer e, that is relatively prime to m.
If e=3, then GCD(3,20)=1. e should be small and prime and so we let e=3.
d is given, d=7.
Public key = (e,n). (Values of e and n are known).
To encrypt a message, we apply the public key to the function 
E(s) = s^e(mod n)

where s is the given message and e and n represent the public key integer pair. In the above question, the plain text M = 5. Plain text needs to be encrypted using the above formula.
   =5³(mod 33)
   = 125 (mod 33)
   = 26
As a result, the encrypted message E(s) = 26. This is what gets transmitted. 

Reference: http://ugcnetsolved-computerscience.blogspot.in/2015/11/rsa-algorithm.html

Answer 2

Answer B) 5

p=3

q=11

z=(p-1)⨉(q-1) =20

Now we know e⨉d=1 mod z  [here d=7]

                    e⨉7 = 1 mod 20

           then,         e=3        

M^' =M^e mod 20

   =5³ mod 20

  =5        

Comments

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Your ans is wrong as it’s M’=Me mod n where n is 33 here .so any will be

M’=5*3 mod33=26

so option D is right