RSA algorithm to encrypt "abcdefghij"

[shivajikobardan]

Say p=5, q=11

z=(p-1)*(q-1)
=40

d relatively prime to z, so d=27

de mod z=1
27e mod 40=1
e=3

$C=M^e mod \; n$

For d, M=4

C=$4^3 mod 55$

Am I right here?

 

[lofgran]

Yes, your calculations are correct. Using the given values of p and q, we can find z to be 40. Since we are looking for a value of d that is relatively prime to z, we can choose d=27. We then solve for e by finding the modular inverse of d mod z, which is 3. Finally, we can use these values to calculate C as $4^3 mod 55$, which is equal to 64 mod 55, or 9. So, your answer is correct.