Part 1)
Ans 1)
Part 1)
Sub groups :
Z7 = {0,1,2,3,4,5,6}
Subgroups of Z7 are :<0>,<1>,<2>,<3>,<4>,<5>,<6>
We know : gcd(7,1)= gcd(7,2)= gcd(7,3)= gcd(7,4) =gcd(7,5) =gcd(7,6)=1
And gcd(7,0)=7
There are 2 subgroups : {0} anf Z7.
Part 2)
Order :
Order is 7 and 1.
Ans 2)
7x + 3y = 5 (mod 11)
4x + 2y = 3 (mod 11)
Hence, 7x+3y+5 and 4x+2y+3 are divisible by 11.
7x + 3y + 5 = 55
4x+2y+3=33
Hence, 7x + 3y = 50
4x + 2y = 30
Hence, x = 5 and y = 5 is a possible solution.
Ans 3)
[ 1 2 3
4 5 6
11 9 8 ]
Determinant = 1 (5×8-9×6)-2(4×8-6×11)+3(4×9-5×11) = (40-54)-2(32-66)+3(36-55)=-14+68-57 = -3
Multiplicative inverse :
= -1/3 [ -14 34 -19
11 -25 13
-3 12 -3]’
= -1/3 [ -14 11 -3
34 -25 12
-19 13 -3]
Ans 4) Euclidean algorithm :
GCD(243,1281) :
1281 / 243 = 5 remainder 66
243 / 66 = 3 remainder 45
66 / 45 = 1 remainder 21
45/21 = 2 remainder 3
21/3= 7 remainder 0
Hence, GCD = 3
243 = 3x3x3x3x3
1281 = 3x7x61
Hence, GCD = 3
Ans 5)
10000010 mod 101101001 = 10000010
Ans 6)
Part 1)
Kerckhoff Principle :
The principle shows that a cryptosystem’s security lies in choosing the keys only. Rest all, even the algorithm is just public knowledge.
Part 2)
Kasiski cryptanalysis :
The Kasiski cryptanalysis is a technique to attack the polyalphabetic substitution cipher like Vigenere cipher.
Ans 7)
1 0 1 0 1
1 0 0 1 1
______
1 0 1 0 1
1 0 1 0 1 x
1 0 1 0 1 x x x x
1 0 1 1 0 1 1 1 1 1
Ans 8)
Part 1) Monoalphabetic cipher vs Polyalphabetic cipher :
Monoalphabetic cipher is used when every symbol in the plain text is mapped with a cipher text’s fixed symbol and Polyalphabetic cipher is the cipher that is based on substitution and uses multiple substitution alphabets.
Part 2) Cryptography vs Steganography :
Cryptography is secret writing and its more popular and Steganography is cover writing and is less popular.
Ans 9) a,c
Sub groups :
Z7 = {0,1,2,3,4,5,6}
Subgroups of Z7 are :<0>,<1>,<2>,<3>,<4>,<5>,<6>
We know : gcd(7,1)= gcd(7,2)= gcd(7,3)= gcd(7,4) =gcd(7,5) =gcd(7,6)=1
And gcd(7,0)=7
There are 2 subgroups : {0} anf Z7.
Part 2)
Order :
Order is 7 and 1.
Ans 2)
7x + 3y = 5 (mod 11)
4x + 2y = 3 (mod 11)
Hence, 7x+3y+5 and 4x+2y+3 are divisible by 11.
7x + 3y + 5 = 55
4x+2y+3=33
Hence, 7x + 3y = 50
4x + 2y = 30
Hence, x = 5 and y = 5 is a possible solution.
Ans 3)
[ 1 2 3
4 5 6
11 9 8 ]
Determinant = 1 (5×8-9×6)-2(4×8-6×11)+3(4×9-5×11) = (40-54)-2(32-66)+3(36-55)=-14+68-57 = -3
Multiplicative inverse :
= -1/3 [ -14 34 -19
11 -25 13
-3 12 -3]’
= -1/3 [ -14 11 -3
34 -25 12
-19 13 -3]
Ans 4) Euclidean algorithm :
GCD(243,1281) :
1281 / 243 = 5 remainder 66
243 / 66 = 3 remainder 45
66 / 45 = 1 remainder 21
45/21 = 2 remainder 3
21/3= 7 remainder 0
Hence, GCD = 3
243 = 3x3x3x3x3
1281 = 3x7x61
Hence, GCD = 3
Ans 5)
10000010 mod 101101001 = 10000010
Ans 6)
Part 1)
Kerckhoff Principle :
The principle shows that a cryptosystem’s security lies in choosing the keys only. Rest all, even the algorithm is just public knowledge.
Part 2)
Kasiski cryptanalysis :
The Kasiski cryptanalysis is a technique to attack the polyalphabetic substitution cipher like Vigenere cipher.
Ans 7)
1 0 1 0 1
1 0 0 1 1
______
1 0 1 0 1
1 0 1 0 1 x
1 0 1 0 1 x x x x
1 0 1 1 0 1 1 1 1 1
Ans 8)
Part 1) Monoalphabetic cipher vs Polyalphabetic cipher :
Monoalphabetic cipher is used when every symbol in the plain text is mapped with a cipher text’s fixed symbol and Polyalphabetic cipher is the cipher that is based on substitution and uses multiple substitution alphabets.
Part 2) Cryptography vs Steganography :
Cryptography is secret writing and its more popular and Steganography is cover writing and is less popular.
Ans 9) a,c
