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Using ‘RSA’ public key cryptosystem, if $p=3$, $q=11$ and $d=7$, find the value of $e$ and encrypt the number $’19’$

  1. $20, 19$
  2. $33,11$
  3. $3, 28$
  4. $77, 28$
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 p=3, q=11 and d=7 

n=pxq= 33

 ϕ(n)=(p-1)x (q-1) =2x10=20

d.e = 1 mod ϕ(n) 

7xe=1 mod 20 or 7e=21 =>e=3

C=P^e mod n

C=19^3 mod33

C=28 

option C is right ans

 

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