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