In 19-24, use the RSA cipher from Examples 8.4.9 and 8.4.10. In 19-21, teranslate the message into its numeric equivalent and encrypt it. In 22-24, decrypt the ciphertext and translate the result in to letters of the alphabet to discover the message.

51 14 49 15

Decrypt the cipher text and write the numeric into letters.

Given information:

The given cipher text is “51 14 49 15”.

Formula used:

Division algorithm:

Let a be an integer and d be a positive integer. Then, there are unique integers q and r with 0≤r<d such that a=dq+r, here, q is the quotient and r is the remainder.

So, q=a div d and r=amodd.

Know that the technique is xm(mody)=[x(mody)]m(mody).

RSA Cipher Work:

The formula of the RSA cryptography is M=Cdmodpq.

To decrypt the numbers, use the formula M=Cdmodpq . …… (1)

Where, d=27 and pq=55.

To decrypt 51, use the formula (1).

M=5127mod55

Its known that, 27 as a sum of powers of 2.

27=24+23+21+20=16+8+2+1

Rewrite the statement as:

M=51( 16+8+2+1)mod55=( 51 16⋅ 518⋅ 512⋅ 511)mod55=( 51 16mod55⋅ 518mod55⋅ 512mod55⋅ 511mod55)mod55=( 51 16mod55⋅ ( 51 4 )2mod55⋅2601mod55⋅51)mod55=( 51 16mod55⋅ ( 51 4 mod55 )2mod55⋅16⋅51)mod55

Simplify the above equation.

( 51 16mod55⋅ ( 51 4 mod55 )2mod55⋅16⋅51)mod55=( 51 16mod55⋅ ( ( 51 2 ) 2 mod55 )2mod55⋅16⋅51)mod55=( 51 16mod55⋅ ( ( 51 2 mod55 ) 2 mod55 )2mod55⋅16⋅51)mod55=( 51 16mod55⋅ ( 16 2 mod55 )2mod55⋅16⋅51)mod55=( ( 51 8 )2mod55⋅ 362mod55⋅16⋅51)mod55

Again, simplify the above equation.

( ( 51 8 )2mod55⋅ 362mod55⋅16⋅51)mod55=( 312mod55⋅31⋅16⋅51)mod55=(961mod55⋅31⋅16⋅51)mod55=(26⋅31⋅16⋅51)mod55=657696mod55=6

Thus, 5127mod55=6.

To decrypt 14, use the formula (1).

M=1427mod55

Its known that, 27 as a sum of powers of 2.

27=24+23+21+20=16+8+2+1

Rewrite the statement as:

M=14( 16+8+2+1)mod55=( 14 16⋅ 148⋅ 142⋅ 141)mod55=( 14 16mod55⋅ 148mod55⋅ 142mod55⋅ 141mod55)mod55=( 14 16mod55⋅ ( 14 4 )2mod55⋅196mod55⋅14)mod55=( 14 16mod55⋅ ( 14 4 mod55 )2mod55⋅31⋅14)mod55

Simplify the above equation.

( 14 16mod55⋅ ( 14 4 mod55 )2mod55⋅31⋅14)mod55=( 14 16mod55⋅ ( ( 14 2 ) 2 mod55 )2mod55⋅31⋅14)mod55=( 14 16mod55⋅ ( ( 14 2 mod55 ) 2 mod55 )2mod55⋅31⋅14)mod55=( 14 16mod55⋅ ( 31 2 mod55 )2mod55⋅31⋅14)mod55=( ( 14 8 )2mod55⋅ 262mod55⋅31⋅14)mod55

Again, simplify the above equation.

( ( 14 8 )2mod55⋅ 262mod55⋅31⋅14)mod55=( 162mod55⋅16⋅31⋅14)mod55=(256mod55⋅16⋅31⋅14)mod55=(36⋅16⋅31⋅14)mod55=249984mod55=9

Thus, 1427mod55=9