UGC NET CSE | June 2013 | Part 2 | Question: 33

Question

Using data p=3, q=11, n=pq, d=7 in RSA algorithm find the cipher text of the given plain text SUZANNE

• A. BUTAEEZ
• B. SUZANNE
• C. XYZABCD
• D. ABCDXYZ

Answer 1

Answer will be (a)

explanation

Encryption RSA algorithm
Cipher text = ((plaintext)^e)mod n
From the given problem we know p =3,q = 11, d =7, n=33 find z :: z = ((p-1)*(q-1)) => 2*10 = 20
Find 'e' or 'd' using the given method : (de) mod z = 1
Therefore (7e)mod 20 = 1 => with trail and error giving e =3, => (7*3)mod 20 = 1
Finding e=3 we go to the encryption by converting the given plain text equivalent number starting a=1,b=2...z=26
Therefore SUZANNE = 19,21,26,1,14,14,5 taking single character and apply encryption
(19^3) mod 33 = 28 ie (28-26) = 2 => B
(21^3) mod 33 = 21 => U
(26^3)mod 33 = 20 => T
(1^3)mod 33 = 1 => A
(14^3)mod 33 = 5 =>E
(5^3)mod 33 = 26 => Z

Comments

but A=65 .. S=83 then how u did  or when use ascii when normal???????

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don't use ascii value in this case,just consider a=1,b=2,c=3 and so on..upto z=26

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r when use ascii when normal???????

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How to calculate mod value for such large numbers