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**Decryption process:**

**Plaintext need to be extracted from the message m.**

13 = m^23 mod 143

n is product of 143

143 is divisible by 11 and 13

So assumption p=11 and q = 13

ϕ(pq)=(11-1)*(13-1) = 120

from modular multiplicative inverse, d = 47

therefore:

plain text = 13^47 mod 143 = 117

So plain text comes to be 117.

__The correctness of decryption
__

To prove correctness of decryption can be done using inverse functions using Chinese Remainder Theorem,

It should prove inverse of decrypt as 13.

M^{ed} = M (mod p) M (mod q) =

M (mod pq) = M (mod n)

117^{23*47} =117(mod 143}

Encryption: c = m^{e} mod n, m < n

117^{23}mod 143 this is equal to 13 hence it is the correct decryption

2.

From input:

p=11, g=2

So p-1 = 10

As both 2 and 11 are co-prime

Now solve 2 ^ 10 mod 11

2^10 mod 11 does equal 1 so 2 has modulo order p-1 where p is a prime and

So hence derived that 2 is a primitive root of 11.

Proved.

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__b__

Given input q =11, n=2, Ya = 9.

Ya = (n^Xa) mod p = (2^Xa) mod 11 = 9

find the power of 2 has modulo so in order that 9 mod 11:

2^1 mod 11 = 2

2^2 mod 11 = 4

2^3 mod 11 = 8

2^4 mod 11 = 5

2^5 mod 11 = 10

2^6 mod 11 = 9

Hence Xa = 6

__c.__

To find the value of K:

K = (Yb ^ Xa) mod q

= 3^6 mod 11

= 3

Value of the secret key is 3.

See our Expert Advice Database Security Assignment

(6/100) Given values in input x = 6 mod 13 and x = 2 mod 17, find x.

Using Chinese Theorem:

In this case:

x ≡ 6 (mod 13) and

x ≡ 2 (mod 17)

17 x_1′ ≡ 1 (mod 13)

13 x_2′ ≡ 1 (mod 17)

To solve 17 x_1′ ≡ 1 (mod 13), note that

17 ≡ 4 (mod 13), hence just solve:

4 x_1′ ≡ 1 (mod 13)

=> x_1′ = 10, x_1 = 170

Solve: 13 x_2′ ≡ 1 (mod 17)

=> x_2′ =4, x_2 =52

Then your (unreduced) result is

x = Σ (a_i * x_i)

= 6 x_1 + 2 x_2

= 6*170 + 2*52 = 1020 + 104

= 1124

Finally we the result modulo for the 13*17:

x = 1124 ≡ 1124 – 5*221 (modulo 221) = 1

So x=19.

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