Solution:
Given,
=>n = 143
=>Plain text(m) = 2
Explanation:
=>n = p*q where p and q are prime number
=>p*q = 143
=>11*13 = 143
=>p = 11 and q = 13
=>(n) = (p)*(q) where (n) is euler totient function of n , (p) is euler totient function of p and (q) is euler totient function of q
=>(n) = (p-1)*(q-1)
=>(n) = (11-1)*(13-1)
=>(n) = 10*12
=>(n) = 120
Finding value of e:
=>Choosing e such that 1<e<(n) and gcd(e,(n)) = 1 where gcd means highest common factor
=>let say e = 11 as it satisfies both the given condition
Finding value of d:
=>Choosing d such that e*dmod(n) = 1
=>11*dmod120 = 1
=>Let say d = 11 as it satifies the above condition
Something special:
=>We have observed that e = d = 11
Finding public key and private key:
=>Pair (e,n) represents the public key
=>Public key = (11,143)
=>Pair (d,n) represents the private key
=>Private key = (11,143)
Finding ciphertext(C):
=>C = m^e mod n
=>C = 2^11 mod 143
=>C = 46
I have explained each and every part with the help of statements attached to it.
(c) In RSA, Given n = 143, compute the public and private keys. Do you observe...
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