Search
Home
Saved
0
0 views
Sign In
Upload
Join
RELATED TITLES
0
quizsols Uploaded by John Doe
Top Charts
Books
Audiobooks
Quiz solutions for a university tutorial
Save
Embed
Share
Print
Lecture2 Updated
1
Download
Magazines
News
of 12
NEST 2014 Question Paper
Encryption
Search document
T U
S
S M S
Documents
Sheet Music
Number Theory and Cryptography
Semest
QuizA — Solutions 1. Find the order of 11 modulo 11 modulo 43 43..
11 modulo 43 must divide 42 by Fermat’s Little Theorem We know that the order of 11 reduces the number of residues that we need to consider. Since 11 7 ≡ 1 (mod 43 ), and for any divisor d of 7 of 7 we we check that 11 d . 1 (mod 1 (mod 43 43), ), the order of 11 modulo 11 modulo 43 43 must must be 7 be 7.. Solution
RSA cryptosy cryptosystem stem has modul modulus us 55 and a decryption exponent 23, what is the encryption ex 2. If an RSA Solution
The encryption exponent exponent is the inverse inverse of 23 of 23 modulo modulo ϕ(55)
=
40, 40 , which is 7 is 7..
3. Suppose you are given two long ciphertexts sct1 and sct2 and told that one of them is some o English text enciphered with a simple substitution cipher and the other is the same Engli enciphered with a Vigenère cipher. If you see the following MAGMA code, which one was (pr enciphered using the Vigenère cipher? > CoincidenceIndex(sct1); 0.04328780874621427836594158797879 > CoincidenceIndex(sct2); 0.0645880574068828150715201231214
Simple substitution ciphers do not change change the coincidence index, index, since they just the frequencies of letters around, whereas Vigenère ciphers tend to lower it because they even frequency frequency distribution distribution of letters. So we can be fairly fairly certain that sct1 is the ciphertext wh enciphered using the Vigen‘ere Vigen‘ere cipher. Solution
4. Find the number of positive integer divisors of 320 of 320.. Solution
Since 320 Since 320
=
2 6 × 51 , we have τ (320)
=
( 6 + 1)(1 + 1)
=
14. 14 .
5. What would be the output of the following MAGMA commands? > V:=VigenereCryptosystem(3); > encipheringkey:=V!“BED”; Enciphering(encipheringkey,Encoding(V,“C Encoding(V,“CORE”)); ORE”)); > Enciphering(encipheringkey,
MasterSolution yourThese semester Scribd commands with tell MAGMA MAGMA to use a Vigenere Vigenere cipher with keywo ke yword rd BED to enci Read Free Foron 30this Days Sign up to vote title an CORE. Since the first letter of the keyword is B which follows A in the alphabet, the first & Theword New York Times Useful Not useful letters should be moved one letter further on in the alphabet; since the second letter of the keyw Special offer for students: Only $4.99/month.
Cancel anytime.
the second letter should be moved four letters further on in the alphabet; since the third lette
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
Upload
Sign In
Read Free For 30 Days Cancel anytime.
Join
Search
Home
Saved
0
0 views
Sign In
Upload
Join
RELATED TITLES
0
quizsols Uploaded by John Doe
Top Charts
Books
Audiobooks
Quiz solutions for a university tutorial
Save
Embed
Share
Print
Lecture2 Updated
1
Download
Magazines
News
of 12
NEST 2014 Question Paper
Encryption
Search document
Number Theory and Cryptography
MATH2
Documents Sheet Music
1 Rearranging this information (or using the extended Euclidean algorithm), we get that 1 19 × 28. 28. So −19 is 19 is an inverse of 28 of 28 modulo modulo 41 41,, and the desired answer is 22 is 22..
=
8. Suppose that an RSA cryptosystem has a public key of ( 55, 3). Encrypt the message [ 2, 4]. encrypt, we raise each letter of the plaintext to the power 3 and reduce modulo 9 (mod 9 (mod 55 55), ), the answer is [ 8, 9].
Solution To 3 ≡
4
=
64
9. What would be the output of the following MAGMA commands? > p:=NextPrime(1024); > n:=p*p*p; > EulerP EulerPhi( hi(n) n) mod p;
These commands commands tell MA MAGMA to let p be the next prime after 1000 (which is 3 1031 , but we don’t need to know that), then let n = p and find the residue of ϕ(n) modulo 3 2 ϕ( p ) = p ( p − 1), the answer is 0 is 0.. Solution
of 34804 modulo 65 modulo 65.. 10. Find the residue of 3
The prime factorization factorization of of 65 65 is 5 × 13, so ϕ(65) = 4 × 12 = 48. Since 4804 ≡ 4 4 ≡ 3 = 81 (mod we have 3 have 3 81 (mod 65 65), ), so the answer is 81 − 65 = 16. 16 . 4804 3 An alternative is to find the residue of 3 modulo the primes 5 and 13 separately, and then c that information by solving a system of simultaneous congruences. Solution
4804
11. If a transposition cipher encrypts the word MELON as ENOML, what is the decryption of RENCO Solution
A transp transposi ositio tion n cipher cipher permut permutes es the letter letterss of the plaint plaintex extt accord according ing to some some fixed fixed per In this case we see that the first letter goes to fourth place, the second letter goes to first place, letter goes to fifth place, fourth letter to third place and fifth letter to second place. So the decry RENCO is CRONE.
of 2291.. 12. Find σ (2291), the sum of all the positive integer divisors of 2291
1, 29, 29, 79, 79, 2291 Since 2291 has prime factorization 29 × 79, its positive divisors are 1, answer is 2400 is 2400.. Solution
Find gcd(4100 − 1, 21). 13. Find gcd Solution
Master Prime yourfactors semester with Scribd 7. So of 21 are 3 and we check check the divisibility divisibility of 4100 1 by each of those Read Free Foron 30this Days Sign up to vote title independently. 3 1 definitely divides 100 therefore, by Fermat Little Theorem, 3 | 4100 & The4New York Times Not useful 6 Useful answer 1 (mod 1 (mod 7) and therefore 4 therefore 4100 4 4 1 (mod 7). Therefore the final is 3. 3 . −
−
≡
≡
.
Cancel anytime.
Special offer for students: Only $4.99/month. Alternatively, one can use the Euler-Fermat theorem straight away.
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
Upload
Sign In
Read Free For 30 Days Cancel anytime.
Join
Search
Home
Saved
0
0 views
Sign In
Upload
Join
RELATED TITLES
0
quizsols Uploaded by John Doe
Top Charts
Books
Audiobooks
Quiz solutions for a university tutorial
Save
Embed
Share
Print
Download
Magazines
News
Lecture2 Updated
1
of 12
NEST 2014 Question Paper
Encryption
Search document
MATH2068/2988
Documents
Hence we have Sheet Music
f (192)
=
f (64) · f (3)
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
=
( 1 − 22 )(1 − 32 ) = 24 .
Read Free Foron 30this Days Sign up to vote title
Not useful Cancel anytime.
Useful
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
Upload
Sign In
Read Free For 30 Days Cancel anytime.
Join
Search
Home
Saved
0
0 views
Sign In
Upload
Join
RELATED TITLES
0
quizsols Uploaded by John Doe
Top Charts
Books
Audiobooks
Quiz solutions for a university tutorial
Save
Embed
Share
Print
Lecture2 Updated
1
Download
Magazines
News
Documents
Sheet Music
of 12
NEST 2014 Question Paper
Encryption
Search document
T U
S
S M S
Number Theory and Cryptography
Semest
QuizB — Solutions 1. Find the order of 2 modulo 2 modulo 23 23..
We know that the order of 2 modulo 23 must divide 22 by Fermat’s Little Theorem reduces the number of residues that we need to consider. Since 2 11 ≡ 1 (mod 23 ), and for any divisor d of 11 of 11 we we check that 2 d . 1 (mod 1 (mod 23 23), ), the order of 2 modulo 2 modulo 23 23 must must be 11 be 11.. Solution
RSA cryptosy cryptosystem stem has modul modulus us 77 and a decryption exponent 43, what is the encryption ex 2. If an RSA Solution
The encryption exponent exponent is the inverse inverse of 43 of 43 modulo modulo ϕ(77)
=
60, 60 , which is 7 is 7..
3. Suppose you are given two long ciphertexts sct1 and sct2 and told that one of them is some o English text enciphered with a simple substitution cipher and the other is the same Engli enciphered with a Vigenère cipher. If you see the following MAGMA code, which one was (pr enciphered using the Vigenère cipher? > CoincidenceIndex(sct1); 0.04328780874621427836594158797879 > CoincidenceIndex(sct2); 0.0645880574068828150715201231214
Simple substitution ciphers do not change change the coincidence index, index, since they just the frequencies of letters around, whereas Vigenère ciphers tend to lower it because they even frequency frequency distribution distribution of letters. So we can be fairly fairly certain that sct1 is the ciphertext wh enciphered using the Vigen‘ere Vigen‘ere cipher. Solution
4. Find the number of positive integer divisors of 200 of 200.. Solution
Since 200 Since 200
=
2 3 × 52 , we have τ (200)
=
( 3 + 1)(2 + 1)
=
12. 12 .
5. What would be the output of the following MAGMA commands? > V:=VigenereCryptosystem(3); > encipheringkey:=V!“BED”; Enciphering(encipheringkey,Encoding(V,“C Encoding(V,“CORE”)); ORE”)); > Enciphering(encipheringkey,
MasterSolution yourThese semester Scribd commands with tell MAGMA MAGMA to use a Vigenere Vigenere cipher with keywo ke yword rd BED to enci Read Free Foron 30this Days Sign up to vote title an CORE. Since the first letter of the keyword is B which follows A in the alphabet, the first & Theword New York Times Useful Not useful letters should be moved one letter further on in the alphabet; since the second letter of the keyw Special offer for students: Only $4.99/month.
Cancel anytime.
the second letter should be moved four letters further on in the alphabet; since the third lette
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
Upload
Sign In
Read Free For 30 Days Cancel anytime.
Join
Search
Home
Saved
0
0 views
Sign In
Upload
Join
RELATED TITLES
0
quizsols Uploaded by John Doe
Top Charts
Books
Audiobooks
Quiz solutions for a university tutorial
Save
Embed
Share
Print
Lecture2 Updated
1
Download
Magazines
News
of 12
NEST 2014 Question Paper
Encryption
Search document
Number Theory and Cryptography
MATH2
Documents Sheet Music
1 Rearranging this information (or using the extended Euclidean algorithm), we get that 1 16 × 30. 30. So −16 is 16 is an inverse of 30 of 30 modulo modulo 37 37,, and the desired answer is 21 is 21..
=
8. Suppose that an RSA cryptosystem has a public key of ( 51, 3). Encrypt the message [ 2, 4]. encrypt, we raise each letter of the plaintext to the power 3 and reduce modulo 13 (mod 13 (mod 51 51), ), the answer is [ 8, 13].
Solution To 3 ≡
4
=
64
9. What would be the output of the following MAGMA commands? > p:=NextPrime(1024); > n:=p*p*p; > EulerP EulerPhi( hi(n) n) mod p;
These commands commands tell MA MAGMA to let p be the next prime after 1000 (which is 3 1031 , but we don’t need to know that), then let n = p and find the residue of ϕ(n) modulo 3 2 ϕ( p ) = p ( p − 1), the answer is 0 is 0.. Solution
of 34004 modulo 55 modulo 55.. 10. Find the residue of 3
The prime factorization factorization of of 55 55 is 5 × 11, so ϕ(55) = 4 × 10 = 40. Since 4004 ≡ 4 4 ≡ 3 = 81 (mod we have 3 have 3 81 (mod 55 55), ), so the answer is 81 − 55 = 26. 26 . 4004 3 An alternative is to find the residue of 3 modulo the primes 5 and 11 separately, and then c that information by solving a system of simultaneous congruences. Solution
4004
11. If a transposition cipher encrypts the word MELON as ENOML, what is the decryption of RENCO Solution
A transp transposi ositio tion n cipher cipher permut permutes es the letter letterss of the plaint plaintex extt accord according ing to some some fixed fixed per In this case we see that the first letter goes to fourth place, the second letter goes to first place, letter goes to fifth place, fourth letter to third place and fifth letter to second place. So the decry RENCO is CRONE.
of 2231.. 12. Find σ (2231), the sum of all the positive integer divisors of 2231
1, 23, 23, 97, 97, 2231 Since 2231 has prime factorization 23 × 97, its positive divisors are 1, answer is 2352 is 2352.. Solution
Find gcd(2100 − 1, 35). 13. Find gcd Solution
Master Prime yourfactors semester with Scribd 7. So of 35 are 5 and we check check the divisibility divisibility of 2100 1 by each of those Read Free Foron 30this Days Sign up to vote title independently. 5 1 definitely divides 100 therefore, by Fermat Little Theorem, 5 | 2100 & The2New York Times Not usefulis 5. 6 Useful answer 1 (mod 1 (mod 7) and therefore 2 therefore 2 100 2 4 1 (mod 1 (mod 7). Therefore the final 5 . −
−
≡
≡
.
Cancel anytime.
Special offer for students: Only $4.99/month. Alternatively, one can use the Euler-Fermat theorem straight away.
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
Upload
Sign In
Read Free For 30 Days Cancel anytime.
Join
Search
Home
Saved
0
0 views
Sign In
Upload
Join
RELATED TITLES
0
quizsols Uploaded by John Doe
Top Charts
Books
Audiobooks
Quiz solutions for a university tutorial
Save
Embed
Share
Print
Download
Magazines
News
Lecture2 Updated
1
of 12
NEST 2014 Question Paper
Encryption
Search document
MATH2068/2988
Documents
Hence we have Sheet Music
f (216)
=
f (8) · f (27)
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
=
( 1 − 22 )(1 − 32 ) = 24 .
Read Free Foron 30this Days Sign up to vote title
Not useful Cancel anytime.
Useful
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
Upload
Sign In
Read Free For 30 Days Cancel anytime.
Join
Search
Home
Saved
0
0 views
Sign In
Upload
Join
RELATED TITLES
0
quizsols Uploaded by John Doe
Top Charts
Books
Audiobooks
Quiz solutions for a university tutorial
Save
Embed
Share
Print
Lecture2 Updated
1
Download
Magazines
News
Documents
Sheet Music
of 12
NEST 2014 Question Paper
Encryption
Search document
T U
S
S M S
Number Theory and Cryptography
Semest
QuizC — Solutions 1. Find the order of 3 modulo 3 modulo 61 61..
We know that the order of 3 modulo 61 must divide 60 by Fermat’s Little Theorem reduces the number of residues that we need to consider. Since 3 10 ≡ 1 (mod 61 ), and for any divisor d of 10 of 10 we we check that 3 d . 1 (mod 1 (mod 61 61), ), the order of 3 modulo 3 modulo 61 61 must must be 10 be 10.. Solution
RSA cryptosy cryptosystem stem has modul modulus us 65 and a decryption exponent 29, what is the encryption ex 2. If an RSA Solution
The encryption exponent exponent is the inverse inverse of 29 of 29 modulo modulo ϕ(65)
=
48, 48 , which is 5 is 5..
3. Suppose you are given two long ciphertexts sct1 and sct2 and told that one of them is some o English text enciphered with a simple substitution cipher and the other is the same Engli enciphered with a Vigenère cipher. If you see the following MAGMA code, which one was (pr enciphered using the Vigenère cipher? > CoincidenceIndex(sct1); 0.04328780874621427836594158797879 > CoincidenceIndex(sct2); 0.0645880574068828150715201231214
Simple substitution ciphers do not change change the coincidence index, index, since they just the frequencies of letters around, whereas Vigenère ciphers tend to lower it because they even frequency frequency distribution distribution of letters. So we can be fairly fairly certain that sct1 is the ciphertext wh enciphered using the Vigen‘ere Vigen‘ere cipher. Solution
4. Find the number of positive integer divisors of 2000 of 2000.. Solution
Since 2000 Since 2000
=
2 4 × 53 , we have τ (2000)
=
( 4 + 1)(3 + 1) = 20. 20 .
5. What would be the output of the following MAGMA commands? > V:=VigenereCryptosystem(3); > encipheringkey:=V!“BED”; Enciphering(encipheringkey,Encoding(V,“C Encoding(V,“CORE”)); ORE”)); > Enciphering(encipheringkey,
MasterSolution yourThese semester Scribd commands with tell MAGMA MAGMA to use a Vigenere Vigenere cipher with keywo ke yword rd BED to enci Read Free Foron 30this Days Sign up to vote title an CORE. Since the first letter of the keyword is B which follows A in the alphabet, the first & Theword New York Times Useful Not useful letters should be moved one letter further on in the alphabet; since the second letter of the keyw Special offer for students: Only $4.99/month.
Cancel anytime.
the second letter should be moved four letters further on in the alphabet; since the third lette
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
Upload
Sign In
Read Free For 30 Days Cancel anytime.
Join
Search
Home
Saved
0
0 views
Sign In
Upload
Join
RELATED TITLES
0
quizsols Uploaded by John Doe
Top Charts
Books
Audiobooks
Quiz solutions for a university tutorial
Save
Embed
Share
Print
Lecture2 Updated
1
Download
Magazines
News
of 12
NEST 2014 Question Paper
Encryption
Search document
Number Theory and Cryptography
MATH2
Documents Sheet Music
1 Rearranging this information (or using the extended Euclidean algorithm), we get that 1 11 × 39. 39. So −11 is 11 is an inverse of 39 of 39 modulo modulo 43 43,, and the desired answer is 32 is 32..
=
8. Suppose that an RSA cryptosystem has a public key of ( 33, 3). Encrypt the message [ 2, 4]. encrypt, we raise each letter of the plaintext to the power 3 and reduce modulo 31 (mod 31 (mod 33 33), ), the answer is [ 8, 31].
Solution To 3 ≡
4
=
64
9. What would be the output of the following MAGMA commands? > p:=NextPrime(1024); > n:=p*p*p; > EulerP EulerPhi( hi(n) n) mod p;
These commands commands tell MA MAGMA to let p be the next prime after 1000 (which is 3 1031 , but we don’t need to know that), then let n = p and find the residue of ϕ(n) modulo 3 2 ϕ( p ) = p ( p − 1), the answer is 0 is 0.. Solution
of 34404 modulo 69 modulo 69.. 10. Find the residue of 3
The prime factorization factorization of of 69 69 is 3 × 23, so ϕ(69) = 2 × 22 = 44. Since 4404 ≡ 4 4 ≡ 3 = 81 (mod we have 3 have 3 81 (mod 69 69), ), so the answer is 81 − 69 = 12. 12 . 4404 3 An alternative is to find the residue of 3 modulo the primes 3 and 23 separately, and then c that information by solving a system of simultaneous congruences. Solution
4404
11. If a transposition cipher encrypts the word MELON as ENOML, what is the decryption of RENCO Solution
A transp transposi ositio tion n cipher cipher permut permutes es the letter letterss of the plaint plaintex extt accord according ing to some some fixed fixed per In this case we see that the first letter goes to fourth place, the second letter goes to first place, letter goes to fifth place, fourth letter to third place and fifth letter to second place. So the decry RENCO is CRONE.
of 2047.. 12. Find σ (2047), the sum of all the positive integer divisors of 2047
1, 23, 23, 89, 89, 2047 Since 2047 has prime factorization 23 × 89, its positive divisors are 1, answer is 2160 is 2160.. Solution
Find gcd(5100 − 1, 21). 13. Find gcd Solution
Master Prime yourfactors semester with Scribd 7. So of 21 are 3 and we check check the divisibility divisibility of 5100 1 by each of those Read Free Foron 30this Days Sign up to vote title independently. 3 1 definitely divides 100 therefore, by Fermat Little Theorem, 3 | 5100 & The5New York Times Not usefulis 3. 6 Useful answer 1 (mod 1 (mod 7) and therefore 5 therefore 5 100 5 4 1 (mod 1 (mod 7). Therefore the final 3 . −
−
≡
≡
.
Cancel anytime.
Special offer for students: Only $4.99/month. Alternatively, one can use the Euler-Fermat theorem straight away.
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
Upload
Sign In
Read Free For 30 Days Cancel anytime.
Join
Search
Home
Saved
0
0 views
Sign In
Upload
Join
RELATED TITLES
0
quizsols Uploaded by John Doe
Top Charts
Books
Audiobooks
Quiz solutions for a university tutorial
Save
Embed
Share
Print
Download
Magazines
News
Lecture2 Updated
1
of 12
NEST 2014 Question Paper
Encryption
Search document
MATH2068/2988
Documents
Hence we have Sheet Music
f (144)
=
f (16) · f (9)
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
=
( 1 − 22 )(1 − 32 ) = 24 .
Read Free Foron 30this Days Sign up to vote title
Not useful Cancel anytime.
Useful
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
Upload
Sign In
Read Free For 30 Days Cancel anytime.
Join
Search
Home
Saved
0
0 views
Sign In
Upload
Join
RELATED TITLES
0
quizsols Uploaded by John Doe
Top Charts
Books
Audiobooks
Quiz solutions for a university tutorial
Save
Embed
Share
Print
Lecture2 Updated
1
Download
Magazines
News
Documents
Sheet Music
of 12
NEST 2014 Question Paper
Encryption
Search document
T U
S
S M S
Number Theory and Cryptography
Semest
QuizD — Solutions 1. Find the order of 3 modulo 3 modulo 41 41..
We know that the order of 3 modulo 41 must divide 40 by Fermat’s Little Theorem reduces the number of residues that we need to consider. Since 3 8 ≡ 1 (mod 41 ), and for any divisor d of 8 of 8 we we check that 3 d . 1 (mod 41 (mod 41), ), the order of 3 modulo 3 modulo 41 41 must must be 8 be 8.. Solution
RSA cryptosy cryptosystem stem has modul modulus us 51 and a decryption exponent 23, what is the encryption ex 2. If an RSA Solution
The encryption exponent exponent is the inverse inverse of 23 of 23 modulo modulo ϕ(51)
=
32, 32 , which is 7 is 7..
3. Suppose you are given two long ciphertexts sct1 and sct2 and told that one of them is some o English text enciphered with a simple substitution cipher and the other is the same Engli enciphered with a Vigenère cipher. If you see the following MAGMA code, which one was (pr enciphered using the Vigenère cipher? > CoincidenceIndex(sct1); 0.04328780874621427836594158797879 > CoincidenceIndex(sct2); 0.0645880574068828150715201231214
Simple substitution ciphers do not change change the coincidence index, index, since they just the frequencies of letters around, whereas Vigenère ciphers tend to lower it because they even frequency frequency distribution distribution of letters. So we can be fairly fairly certain that sct1 is the ciphertext wh enciphered using the Vigen‘ere Vigen‘ere cipher. Solution
4. Find the number of positive integer divisors of 400 of 400.. Solution
Since 400 Since 400
=
2 4 × 52 , we have τ (400)
=
( 4 + 1)(2 + 1)
=
15. 15 .
5. What would be the output of the following MAGMA commands? > V:=VigenereCryptosystem(3); > encipheringkey:=V!“BED”; Enciphering(encipheringkey,Encoding(V,“C Encoding(V,“CORE”)); ORE”)); > Enciphering(encipheringkey,
MasterSolution yourThese semester Scribd commands with tell MAGMA MAGMA to use a Vigenere Vigenere cipher with keywo ke yword rd BED to enci Read Free Foron 30this Days Sign up to vote title an CORE. Since the first letter of the keyword is B which follows A in the alphabet, the first & Theword New York Times Useful Not useful letters should be moved one letter further on in the alphabet; since the second letter of the keyw Special offer for students: Only $4.99/month.
Cancel anytime.
the second letter should be moved four letters further on in the alphabet; since the third lette
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
Upload
Sign In
Read Free For 30 Days Cancel anytime.
Join
Search
Home
Saved
0
0 views
Sign In
Upload
Join
RELATED TITLES
0
quizsols Uploaded by John Doe
Top Charts
Books
Audiobooks
Quiz solutions for a university tutorial
Save
Embed
Share
Print
Lecture2 Updated
1
Download
Magazines
News
of 12
NEST 2014 Question Paper
Encryption
Search document
Number Theory and Cryptography
MATH2
Documents Sheet Music
1 Rearranging this information (or using the extended Euclidean algorithm), we get that 1 21 × 38. 38. So −21 is 21 is an inverse of 38 of 38 modulo modulo 47 47,, and the desired answer is 26. 26.
=
8. Suppose that an RSA cryptosystem has a public key of ( 46, 3). Encrypt the message [ 2, 4]. encrypt, we raise each letter of the plaintext to the power 3 and reduce modulo 18 (mod 18 (mod 46 46), ), the answer is [ 8, 18].
Solution To 3 ≡
4
=
64
9. What would be the output of the following MAGMA commands? > p:=NextPrime(1024); > n:=p*p*p; > EulerP EulerPhi( hi(n) n) mod p;
These commands commands tell MA MAGMA to let p be the next prime after 1000 (which is 3 1031 , but we don’t need to know that), then let n = p and find the residue of ϕ(n) modulo 3 2 ϕ( p ) = p ( p − 1), the answer is 0 is 0.. Solution
of 36004 modulo 77 modulo 77.. 10. Find the residue of 3
The prime factorization factorization of of 77 77 is 7 × 11, so ϕ(77) = 6 × 10 = 60. Since 6004 ≡ 4 4 ≡ 3 = 81 (mod we have 3 have 3 81 (mod 77 77), ), so the answer is 81 − 77 = 4. 4 . 6004 3 An alternative is to find the residue of 3 modulo the primes 7 and 11 separately, and then c that information by solving a system of simultaneous congruences. Solution
6004
11. If a transposition cipher encrypts the word MELON as ENOML, what is the decryption of RENCO Solution
A transp transposi ositio tion n cipher cipher permut permutes es the letter letterss of the plaint plaintex extt accord according ing to some some fixed fixed per In this case we see that the first letter goes to fourth place, the second letter goes to first place, letter goes to fifth place, fourth letter to third place and fifth letter to second place. So the decry RENCO is CRONE.
of 2407.. 12. Find σ (2407), the sum of all the positive integer divisors of 2407
1, 29, 29, 83, 83, 2407 Since 2407 has prime factorization 29 × 83, its positive divisors are 1, answer is 2520 is 2520.. Solution
Find gcd(3100 − 1, 35). 13. Find gcd Solution
Master Prime yourfactors semester with Scribd 7. So of 35 are 5 and we check check the divisibility divisibility of 3100 1 by each of those Read Free Foron 30this Days Sign up to vote title independently. 5 1 definitely divides 100 therefore, by Fermat Little Theorem, 5 | 3100 & The3New York Times Not usefulis 5. 6 Useful answer 1 (mod 1 (mod 7) and therefore 3 therefore 3 100 3 4 1 (mod 1 (mod 7). Therefore the final 5 . −
−
≡
≡
.
Cancel anytime.
Special offer for students: Only $4.99/month. Alternatively, one can use the Euler-Fermat theorem straight away.
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
Upload
Sign In
Read Free For 30 Days Cancel anytime.
Join
Search
Home
Saved
0
0 views
Sign In
Upload
Join
RELATED TITLES
0
quizsols Uploaded by John Doe
Top Charts
Books
Audiobooks
Quiz solutions for a university tutorial
Save
Embed
Share
Print
Download
Magazines
News
Lecture2 Updated
1
of 12
NEST 2014 Question Paper
Encryption
Search document
MATH2068/2988
Documents
Hence we have Sheet Music
f (324)
=
f (4) · f (81)
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
=
( 1 − 22 )(1 − 32 ) = 24 .
Read Free Foron 30this Days Sign up to vote title
Not useful Cancel anytime.
Useful
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.
Upload
Sign In
Read Free For 30 Days Cancel anytime.
Join
