CS 472
Network and Systems Security
Fall 2005
Final Exam
Time 2 & 1/2 hours
Open Book & Notes
Name:
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Each question is 10 points.
1
1. Assuming we are using mod 13 arithmetic. Let K = 5. What is the value of:
 Additive inverse of K.
The inverse is 8, since 5+8 = 0 mod 13
 Multiplicative inverse of K.
The inverse is 8, since 5x8 = 1 mod 13
 Exponentiative inverse of K.
Since 13 is prime, phi (13) = 12
Thus the inverse is 5, since 5x5 = 1 mod 12
2. In the above question, let C = 3 be a cipher of a message M, what is the value of
M if it were encrypted using:
 the Additive K.
M = 11 since 3+8 = 11 mod 13
 the Multiplicative K.
M = 11 since 3x8 = 11 mod 13
 the Exponentiative K.
M = 9 since 3**5 mod 13 = 9
2
3. In RSA, assume e=3, p = 11 and q =23.
Show that 147 is a possible value of d.
n = pxq = 11x23 = 253
phid (n) = (p-1)x(q-1) = (10)x(22) = 220
we need to show that : exd= 1 phi(n)
3x147 = 441 = 1 mod 220
Thus d= 147 is the inverse of e=3.
4. Calculate the value of 857 mod 100 without using a calculator.
57 =
1
8**57=
=
=
=
=
=
1
8**2)*8
12
1
**2)*8
0
0
**2)
**2)
1
**2)*8
…..
52 ……
4 …….
16 ……
48
3
5. Consider Diffie-Hellman with p=7 and g=5. Assume Alice picket 6 as her
random number while Bob picked 8 as his random number. What is the value of
the shared secret between Alice and Bob following the Diffie-Hellman message
exchange?
SA = 6
SB = 8
PA = 5**SA mod 7 = 5**6 mod 7 = 1
PB = 5**SB mod 7 = 5**8 mod 7 = 4
SA= PB**SA mod 7 = 4**6 mod 7 = 1
SB = PA**SB mod 7 = 1**8 mod 7 = 1
4
6. Assume a UNIX server has a password file containing the hash of 3,000 user
passwords. Assume that a person has access to dictionary containing 2,000,000
possible passwords. How many hash operations are required to perform a
dictionary attack on such password file if:
a. The server is not using a salt.
The size of dictionary: 2,000,000
b. The server is using a salt.
The size of dictionary * the size of the password file =
2,000,000 * 3,000 = 6x10**9
5
7. In sending signed-encrypted mail using openssl SMIME explain why the
receiver should have access to:
i. The receiver’s private key.
So the receiver can get the symmetric key for decrypting the message
ii. The sender’s certificate.
So the receiver can verify the signature of the sender on the message digest
and get assurance of the identity of the sender and that the message is not
altered.
6
8. The following is a mutual authentication protocol:
Alice
M1= I'm Alice, K{timestamp T}
<
Bob
>
M2 = K{T ++}
 Explain why:
o Bob is sure that the sender of M1 is Alice.
Since he can check if the timestamp is roughly equals to his.
o Alice is sure that the sender of M2 is Bob.
She can check that M2 value is 1 + the time sent in M1.
9. In the above question, what is the pitfall of replacing the timestamp T with a
random number R.
In such case Bob can not verify the identity of Alice.
7
10. The following are the listings of certificate request and an issued certificate.
Certificate Request
Data:
Version: 0 (0x0)
Subject: C=US, ST=Virginia, L=Norfolk, O=Old Dominion University, OU=Computer
Science Department, CN=Charles A. Morris/emailAddress=cmorris@cs.odu.edu
Subject Public Key Info:
Public Key Algorithm: rsaEncryption
RSA Public Key: (1024 bit)
Modulus (1024 bit):
00:b6:08:6e:90:ca:ba:5c:b7:83:e2:d6:64:6f:33:
26:f9:33:c2:d2:56:ab:3e:f2:9a:68:87:b3:53:06:
eb:0b:2e:00:4d:d6:8e:15:5a:eb:32:dc:b3:1e:a8:
9e:6c:c3:26:dc:ef:2c:82:4b:32:05:8b:88:73:8a:
91:91:71:c8:41:a3:5a:c8:12:6f:19:d7:0d:76:8e:
21:ed:e5:b1:ea:69:1e:f9:f3:65:c6:0d:33:e3:61:
aa:31:41:1c:53:85:a8:d3:71:b4:7d:96:48:fc:2e:
5e:df:ed:54:b7:c5:fd:ce:80:eb:8b:2e:c3:f1:17:
65:fa:85:31:5e:b5:3c:74:6b
Exponent: 65537 (0x10001)
Attributes:
challengePassword
:canttouchthis
Signature Algorithm: md5WithRSAEncryption
1e:cb:9e:0c:f4:01:2e:a6:4f:c4:0d:0e:55:49:9b:ca:a3:2b:
bb:5d:c7:ea:29:91:a8:37:5e:c3:23:cf:81:6a:c4:1a:13:6c:
4a:1b:be:59:7b:6a:c6:80:6b:ba:50:31:76:67:b9:f9:67:e6:
fa:2c:fe:6e:a3:19:f1:2a:86:aa:64:72:a7:5c:c3:76:9d:38:
55:f2:8c:c0:f8:1a:41:4c:c2:71:a2:29:28:2e:14:5b:d5:fc:
bf:3d:cf:5f:b6:98:40:9d:d1:1c:e8:97:f2:4e:41:9b:4b:8b:
10:c0:6a:0b:dc:f8:e1:3c:bb:01:84:64:8d:43:24:17:42:15:
d8:83
8
Data:
Version: 3 (0x2)
Serial Number: 2 (0x2)
Signature Algorithm: md5WithRSAEncryption
Issuer: CN=Dr. Wahab, ST=Virginia, C=US/emailAddress=wahab@cs.odu.edu, O=Old
Dominion University
Validity
Not Before: Oct 26 20:03:16 2005 GMT
Not After : Oct 26 20:03:16 2006 GMT
Subject: CN=Charles A. Morris, ST=Virginia, C=US/emailAddress=cmorris@cs.odu.edu, O=Old
Dominion University, OU=Computer Science Department
Subject Public Key Info:
Public Key Algorithm: rsaEncryption
RSA Public Key: (1024 bit)
Modulus (1024 bit):
00:b6:08:6e:90:ca:ba:5c:b7:83:e2:d6:64:6f:33:
26:f9:33:c2:d2:56:ab:3e:f2:9a:68:87:b3:53:06:
eb:0b:2e:00:4d:d6:8e:15:5a:eb:32:dc:b3:1e:a8:
9e:6c:c3:26:dc:ef:2c:82:4b:32:05:8b:88:73:8a:
91:91:71:c8:41:a3:5a:c8:12:6f:19:d7:0d:76:8e:
21:ed:e5:b1:ea:69:1e:f9:f3:65:c6:0d:33:e3:61:
aa:31:41:1c:53:85:a8:d3:71:b4:7d:96:48:fc:2e:
5e:df:ed:54:b7:c5:fd:ce:80:eb:8b:2e:c3:f1:17:
65:fa:85:31:5e:b5:3c:74:6b
Exponent: 65537 (0x10001)
X509v3 extensions:
X509v3 Basic Constraints:
CA:FALSE
Signature Algorithm: md5WithRSAEncryption
1c:17:53:99:21:c7:07:9f:b2:77:b5:0a:18:23:e6:73:62:60:
a8:7d:18:26:75:4a:a8:8c:09:ff:c9:00:02:c6:21:0e:93:82:
ac:05:8a:1d:46:81:8a:6e:11:d2:f4:34:9f:f7:bf:7e:30:6d:
c7:62:b5:dc:86:81:1e:ee:e9:ac:e9:d2:2e:70:f6:ef:a7:f4:
60:69:25:ca:f2:7f:4c:43:fb:29:62:dd:09:1c:25:0f:b6:59:
32:3e:65:c8:0d:07:6f:ae:cd:0f:d1:d2:3c:7a:1d:a2:f3:81:
93:32:3b:d4:f0:8b:7d:95:b0:a1:b5:f5:c2:9e:f1:27:1a:f7:
ad:51
Issued Certificate:
9
a. Who is the requester?
cmorris
b. Who is the Certificate Authority?
Dr. wahab
c. What is the value of e of the public key <e, n>?
65537
d. What is the length of d of the private key <d,n>?
1024 bits
e. What is the value of the first five octets of n of the public key <e, n>?
00:b6:08:6c:90
f. Why the signatures on the request and on the certificate are different?
The signature on the request is cmorris while the signature on the certificate is
Dr. wahab
10