Given

mod n = 221

public key e = 7

To find = $p=?$ $q=?$

$\phi(n)=?$ $d=?$

Using RSA

Also Encrypt M = 5

Solution =

n = p x q

221 = 13 X 17

P = 13

Q = 17

a. $\phi(n)=(p-1)(q-1)$

= (13-1)(17-1)

= 12 X 16

= 192

Now, d = ?

b. d x e mod (p-1)(q-1) = 1

i.e.

d x e mod(p-1)(q-1) = 1

e = 07

p = 13

q = 17

$\therefore$ d x 7 mod(31-1)(17-1) = 1

$\therefore$ d x 7 mod192 = 1

$\therefore$ d = 55

c. Encryption : M = 5

C = M^e mod n

i.e.

C = 5⁷ mod 221

C = 112

d. Decryption :

P = ( CT)^d mod n

= (112)⁵⁵ mod 221

P = 5

Thus,

P = 13

Q = 17

$\phi(n)=192$

d = 55

Encrypted for M = 5

C = 112