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,
(n + 2^2 )mod10 where n =0, 1, 2,...,9 , without being intercepted and decrypted by other countries. Using number theory method:
(a) State the decrypted message.
(b) Determine the decryption key.
(c) Suggest an improvement to the encryption key to increase the encryption strength

Answer 1

Answer on Question #63412 – Math – Discrete Mathematics

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,

$$(n + 2^2) \text{mod}10$$

where $n = 0,1,2,\ldots,9$,

without being intercepted and decrypted by other countries. Using number theory method:

Question

(a) State the decrypted message.

Solution

(a)

We shall encrypt the message "161803398" using Caesar cipher with the encryption key $(n + 2^2) \text{mod}10$:

$n = 1$: $y = (1 + 4) \text{mod}10 = 5$, $y$ is a symbol of the encrypted message;

$n = 6$: $y = (6 + 4) \text{mod}10 = 0$;

$n = 1$: $y = (1 + 4) \text{mod}10 = 5$;

$n = 8$: $y = (8 + 4) \text{mod}10 = 2$;

$n = 0$: $y = (0 + 4) \text{mod}10 = 4$;

$n = 3$: $y = (3 + 4) \text{mod}10 = 7$;

$n = 3$: $y = (3 + 4) \text{mod}10 = 7$;

$n = 9$: $y = (9 + 4) \text{mod}10 = 3$;

$n = 8$: $y = (8 + 4) \text{mod}10 = 2$.

Answer:

Encrypted message is 505247732.

Question

(b) Determine the decryption key.

Solution

(b)

Decryption key:

$$x = (y - 4 + 10) \mod 10;$$

$x = (y + 6) \mod 10$, where $x$ is a symbol of the original message.

Answer: decryption key is $x = (y + 6) \mod 10$.

Question

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

Solution

(c) Example of improvement to the encryption key:

$$(n + 30) \mod 20$$

The encryption strength of this key is greater because every symbol of original message is encrypted with two symbols.

Answer: $(n + 30) \mod 20$.

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