An RSA cipher has public key pq = 65 and e = 7.
(a) Translate the message YES into its numeric equivalent, and use the formula $C = M ^ { e }$ (mod pq) to encrypt the message.
(b) Decrypt the ciphertext 50 16 and translate the result into letters of the alphabet to discover the message.

Question

An RSA cipher has public key pq = 65 and e = 7.
(a) Translate the message YES into its numeric equivalent, and use the formula $C = M ^ { e }$  (mod pq) to encrypt the message.
(b) Decrypt the ciphertext 50 16 and translate the result into letters of the alphabet to discover the message.

Quizplus · Accepted Answer

The Answer is a. The numeric equivalents of

\mathrm { Y } , \mathrm { E }

, and

\mathrm { S }

are 25,5 , and 19 . So to encrypt these letters, the following quantities must be computed:

25 ^ { 7 } \bmod 65,5 ^ { 7 } \bmod 65

, and

19 ^ { 7 } \bmod 65

.
Y:

25 ^ { 7 } \bmod 65 = 25 \quad

E:

5 ^ { 7 } \bmod 65 = 60 \quad

S:

19 ^ { 7 } \bmod 65 = 59

Therefore, the encrypted message is 25 60 59 .
b. Because

p q = 65 = 5 \cdot 13

and because both 5 and 13 are prime numbers,

( p - 1 ) ( q - 1 ) =

4 \cdot 12 = 48

. The decryption key

d

is the solution to the congruence

7 d \equiv 1 ( \bmod 48 )

. Using either trial-and-error or the solution to problem 18 gives that

d = 7

. So to decrypt the message, the following quantities must be computed:

50 ^ { 7 } \bmod 65

, and

41 ^ { 7 } \bmod 65

.
Since

50 ^ { 7 } \bmod 65 = 15

and

16 ^ { 7 } \bmod 65 = 16

, and since 15 corresponds to

\mathrm { N }

and 16 corresponds to

\mathrm { O }

the decrypted message is NO .

An RSA Cipher Has Public Key Pq = 65 and E