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What will be the cipher text produced by the following cipher function for the plain text ISRO with key k =7. [Consider 'A' = 0, 'B' = 1, .......'Z' = 25]. Ck(M) = (kM + 13) mod 26
- a)RJCH
- b)QIBG
- c)GQPM
- d)XPIN
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
What will be the cipher text produced by the following cipher function...
Ck(M) = (kM + 13) mod 26
Here, 'A' = 0, 'B' = 1, .......'Z' = 25
I = Ck(I) = (7*8 + 13) mod 26 = 17 = R
S = Ck(S) = (7*18 + 13) mod 26 = 9 = J
R = Ck(R) = (7*17 + 13) mod 26 = 2 = C
O = Ck(O) = (7*14 + 13) mod 26 = 7 = H
So, option (A) is correct.
S = Ck(S) = (7*18 + 13) mod 26 = 9 = J
R = Ck(R) = (7*17 + 13) mod 26 = 2 = C
O = Ck(O) = (7*14 + 13) mod 26 = 7 = H
So, option (A) is correct.
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What will be the cipher text produced by the following cipher function...
Solution:
Given, plain text = ISRO
Key, k = 7
A = 0, B = 1, .......Z = 25
Cipher function: Ck(M) = (kM 13) mod 26
To find the cipher text, we need to substitute the plain text alphabets one by one in the cipher function and apply the given key.
For I, Ck(M) = (kM 13) mod 26
Substituting M = 8 (as I is the 9th alphabet), k = 7, we get
Ck(8) = (7*8 + 13) mod 26
Ck(8) = 69 mod 26
Ck(8) = 17
So, I is encrypted to 17, which is R.
Similarly, for S, Ck(M) = (kM 13) mod 26
Substituting M = 18 (as S is the 19th alphabet), k = 7, we get
Ck(18) = (7*18 + 13) mod 26
Ck(18) = 139 mod 26
Ck(18) = 11
So, S is encrypted to 11, which is J.
For R, Ck(M) = (kM 13) mod 26
Substituting M = 17 (as R is the 18th alphabet), k = 7, we get
Ck(17) = (7*17 + 13) mod 26
Ck(17) = 128 mod 26
Ck(17) = 24
So, R is encrypted to 24, which is C.
For O, Ck(M) = (kM 13) mod 26
Substituting M = 14 (as O is the 15th alphabet), k = 7, we get
Ck(14) = (7*14 + 13) mod 26
Ck(14) = 111 mod 26
Ck(14) = 9
So, O is encrypted to 9, which is H.
Therefore, the cipher text for ISRO with key k=7 is RJCH.
Hence, option A is the correct answer.
Given, plain text = ISRO
Key, k = 7
A = 0, B = 1, .......Z = 25
Cipher function: Ck(M) = (kM 13) mod 26
To find the cipher text, we need to substitute the plain text alphabets one by one in the cipher function and apply the given key.
For I, Ck(M) = (kM 13) mod 26
Substituting M = 8 (as I is the 9th alphabet), k = 7, we get
Ck(8) = (7*8 + 13) mod 26
Ck(8) = 69 mod 26
Ck(8) = 17
So, I is encrypted to 17, which is R.
Similarly, for S, Ck(M) = (kM 13) mod 26
Substituting M = 18 (as S is the 19th alphabet), k = 7, we get
Ck(18) = (7*18 + 13) mod 26
Ck(18) = 139 mod 26
Ck(18) = 11
So, S is encrypted to 11, which is J.
For R, Ck(M) = (kM 13) mod 26
Substituting M = 17 (as R is the 18th alphabet), k = 7, we get
Ck(17) = (7*17 + 13) mod 26
Ck(17) = 128 mod 26
Ck(17) = 24
So, R is encrypted to 24, which is C.
For O, Ck(M) = (kM 13) mod 26
Substituting M = 14 (as O is the 15th alphabet), k = 7, we get
Ck(14) = (7*14 + 13) mod 26
Ck(14) = 111 mod 26
Ck(14) = 9
So, O is encrypted to 9, which is H.
Therefore, the cipher text for ISRO with key k=7 is RJCH.
Hence, option A is the correct answer.
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