Respuesta :
Solution :
It is given in the question that :
Two prime numbers : p = 41
q = 17
Therefore, n = p x q
= 41 x 17
= 697
Now we know that :
[tex]$\phi (n) = (p-1)\times (q-1)$[/tex]
[tex]$\phi (n) = (41-1)\times (17-1)$[/tex]
= 40 x 16
= 640
a). For the public key [tex]$\text{e gcd(e, }\phi(n))=1$[/tex]
Now the [tex]$\text{gcd}(e_1,\phi(n))=1$[/tex] implies [tex]$\text{gcd}(32,640)=1$[/tex]
But this is false as the [tex]$\text{gcd}(32,640)! = 1$[/tex]
Therefore the public key will prefer [tex]$e_2$[/tex] , that is 49.
b). We have to find the private key d. So we know that
[tex]$e \times d = 1 \text{ mod } \phi(n)$[/tex]
[tex]$49 \times d = 1 \text{ mod } 640$[/tex]
Therefore the value of d = 209
c). The encryption of the message M, we will use the relation:
[tex]$C=M^e \text{ mod } n$[/tex] ; here "C" is cipher text
Given M = 26 and we know that [tex]$e=49$[/tex] and [tex]$n=697$[/tex]
Therefore, [tex]$C=26^{49} \text{ mod }697$[/tex]
= 468
Thus the cipher text for the plain text 26 is 468.
d). For the decryption of message C,
[tex]$M=C^d \text{ mod }n$[/tex] (here C = cipher text, M = plain text)
Given [tex]$ C=513$[/tex]. And the value of d is 209 and n is 697
Therefore, [tex]$M=513^{209}\text{ mod } 697$[/tex]
[tex]$=326$[/tex]
Therefore, the plain text for cipher text 513 is 326.
