Question #63939

Suppose that you need to deliver the message “161803398” which is the pass key for a weapon
activation to country X. Encrypt the message using Caesar cipher with the encryption key,
2
( 2 )mod9 n 
where
n 1, 2,...,9 , without being intercepted and decrypted by other countries.
(a) State the decrypted message.
(b) Determine the decryption key.
(c) Suggest an improvement to the encryption key to increase the encryption strength

Expert's answer

Answer on Question #63939 - Math - Discrete Mathematics

Question

Suppose that you need to deliver the message "161803398" which is the pass key for a weapon activation to country X. Encrypt the message using Caesar cipher with the encryption key,

2

(2) mod 9 n +

where

n=1,2,,9n = 1,2,\ldots ,9 , without being intercepted and decrypted by other countries.

(a) State the decrypted message.

(b) Determine the decryption key.

(c) Suggest an improvement to the encryption key to increase the encryption strength

Remarks:

Note the only variant of the correct (original) condition is possible if the distorted text

2

(2) mod 9 n +

where

n=1,2,,9n = 1,2,\ldots ,9

is replaced by

(n+22)(n + 2\cdot 2) mod 10, where n=0,1,2,,9n = 0,1,2,\ldots ,9

and

'(a) State the decrypted message'

is replaced by

'(a) State the encrypted message'.

Solution

(a) Each digit dd of the encrypted message is calculated as d=(n+4)mod10d = (n + 4) \mod 10 , where nn is the digit of the message to be encrypted, in the same position. Thus

1(1+4)mod10=51\to (1 + 4)\mod 10 = 5

6(6+4)mod10=06\to (6 + 4)\mod 10 = 0

1(1+4)mod10=51\to (1 + 4)\mod 10 = 5

8(8+4)mod10=28\to (8 + 4)\mod 10 = 2

0(0+4)mod10=40\to (0 + 4)\mod 10 = 4

3(3+4)mod10=73\to (3 + 4)\mod 10 = 7

3(3+4)mod10=73\to (3 + 4)\mod 10 = 7

9(9+4)mod10=39\to (9 + 4)\mod 10 = 3

8(8+4)mod10=28\to (8 + 4)\mod 10 = 2

and the encrypted message is "505247732".

(b) From the known digit dd of the encrypted message, the corresponding digit nn of the decrypted message can be found as n=(n+44+10)mod10=(d+104)mod10=n = (n + 4 - 4 + 10) \mod 10 = (d + 10 - 4) \mod 10 =

=(d+6)mod10= (d + 6) \mod 10 .

(c) An example of the improvement to the encryption key is

d=(yn+x)mod10,d = (y\cdot n + x)\mod 10,

where n=0,1,2,,9n = 0,1,2,\ldots ,9 is the digit of the message to be encrypted,

d=0,1,2,,9d = 0,1,2,\ldots ,9 is the digit of the encrypted message,

the pair (x,y)(x, y) is the encryption key, y=1,3,7,9y = 1,3,7,9 (each yy is such that yy and 10 have no common divisors), x=0,1,2,,9x = 0,1,2,\ldots,9, if y=1y = 1, x0x \neq 0 (assuming that each digit of the encrypted message has to be different from the corresponding digit of the decrypted message).

Thus, there exist 10·4-1=39 variants of the key.

Let v=y3mod10v = y^3 \mod 10, u=v(10x)mod10u = v(10 - x) \mod 10. Then (vx+u)mod10=0(v \cdot x + u) \mod 10 = 0,

vy=y4mod10=1v \cdot y = y^4 \mod 10 = 1 (vv is reverse yy modulo 10).

Consequently,


(vd+u)mod10=(vyn+vx+u)mod10=nmod10=n.(v \cdot d + u) \mod 10 = (v \cdot y \cdot n + v \cdot x + u) \mod 10 = n \mod 10 = n.


The other example of the improvement to the encryption key is


di=(ni+xi)mod10,d_i = (n_i + x_i) \mod 10,


where

i=1,2,3,9i = 1,2,3,\ldots 9 is a position of the digit of the message,

di=0,1,2,9d_i = 0,1,2,\ldots 9 is the digit of the encrypted message at the position ii,

ni=0,1,2,9n_i = 0,1,2,\ldots 9 is the digit of the message to be encrypted, at the position ii,

xi=1,2,9x_i = 1,2,\ldots 9 (assuming that each digit of the encrypted message has to be different from the corresponding digit of the decrypted message).

Thus, the encryption key is a sequence {x1,x2,x3,x4,x5,x6,x7,x8,x9}\{x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9\}

This sequence can take 99=3874204899^9 = 387420489 values.

The decryption is ni=(di+10xi)mod10n_i = (d_i + 10 - x_i) \mod 10

Answer:

(a) The encrypted message is "505247732".

(b) The decryption key is (d+6)mod10(d + 6) \mod 10, where d=0,1,2,,9d = 0, 1, 2, \ldots, 9

(c) The improvement to the encryption key:

Example 1

Encryption: d=(yn+x)mod10d = (y \cdot n + x) \mod 10, where n=0,1,2,,9n = 0,1,2,\ldots,9 for each digit.

y=1,3,7,9y = 1,3,7,9, x=0,1,2,,9x = 0,1,2,\ldots,9, if y=1y = 1, x0x \neq 0. The pair (x,y)(x,y) is the encryption key, there exist 39 variants of the key.

Decryption: (vd+u)mod10(v \cdot d + u) \mod 10, where v=y3mod10v = y^3 \mod 10, u=v(10x)mod10u = v(10 - x) \mod 10.

Example 2

Encryption: di=(ni+xi)mod10d_i = (n_i + x_i) \mod 10, where i=1,2,3,9i = 1,2,3,\ldots 9 is a position of the digit of the message, di=0,1,2,9d_i = 0,1,2,\ldots 9 is the digit of the encrypted message at the position ii, ni=0,1,2,9n_i = 0,1,2,\ldots 9 is the digit of the message to be encrypted, at the position ii, xi=1,2,9x_i = 1,2,\ldots 9. The encryption key is a sequence {x1,x2,x3,x4,x5,x6,x7,x8,x9}\{x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9\}, it takes 99=9^9 =

=387420489= 387420489 variants.

Decryption: ni=(di+10xi)mod10n_i = (d_i + 10 - x_i) \mod 10.

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