   Chapter 8.4, Problem 22ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# 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. 13 20 20 09

To determine

Decrypt the cipher text and write the numeric into letters.

Explanation

Given information:

The given cipher text is “13 20 20 09”.

Formula used:

Division algorithm:

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

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 13, use the formula (1).

M=1327mod55

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=13( 16+8+2+1)mod55=( 13 16 138 132 131)mod55=( 13 16mod55 138mod55 132mod55 131mod55)mod55=( 13 16mod55 ( 13 4 )2mod55169mod5513)mod55=( 13 16mod55 ( 13 4 mod55 )2mod55413)mod55

Simplify the above equation.

( 13 16mod55 ( 13 4 mod55 )2mod55413)mod55=( 13 16mod55 ( ( 13 2 ) 2 mod55 )2mod55413)mod55=( 13 16mod55 ( ( 13 2 mod55 ) 2 mod55 )2mod55413)mod55=( 13 16mod55 ( 4 2 mod55 )2mod55413)mod55=( ( 13 8 )2mod55 162mod55413)mod55

Again, simplify the above equation.

( ( 13 8 )2mod55 162mod55413)mod55=( 362mod5536413)mod55=(1296mod5536413)mod55=(3136413)mod55=58032mod55=7

Thus, 1327mod55=7.

To decrypt 20, use the formula (1).

M=2027mod55

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=20( 16+8+2+1)mod55=( 20 16 208 202 201)mod55=( 20 16mod55 208mod55 202mod55 201mod55)mod55=( 20 16mod55 ( 20 4 )2mod55400mod5520)mod55=( 20 16mod55 ( 20 4 mod55 )2mod551520)mod55

Simplify the above equation

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