1. Computer and Network Security.
Name: Khalid Khalil Kamil
Matric. No.: G0327887.
Solutions to Assignment # 3:
1- Use Euclid’s algorithm to find the greatest common divisor (gcd) of:
a. gcd(14,105):
Solution:
gcd(105,14)=gcd(14, 105 mod 14) = gcd(14, 7) = gcd(7,14 mod 7) = gcd(7,0)=7.
b. gcd(180, 1512)
Solution:
Gcd(1512,180)=gcd(180,1512 mod180)=gcd(180,72)=gcd(72,36)=gcd(36,72mod36)
=gcd(36,0)= 36.
c. gcd(1001,7655):
solution:
gcd(7655,1001)=gcd(1001,7655mod1001)=gcd(1001,648)=gcd(648,1001mod648)
=gcd(648,353)=gcd(353,648mod353)=gcd(353,295)=gcd(295,353mod295) =gcd(295,58)
=gcd(58,295mod58)=gcd(58,5)=gcd(5,58mod5)=gcd(5,3)=gcd(3,5mod3)=gcd(3,2)
=gcd(2,3mod2)=gcd(2,1)=gcd(1,2mod1)=gcd(1,0)=1.
d. gcd(24140,16762):
solution:
gcd(24140,16762)=gcd(16762,24140mod16762)=gcd(16762,7378)
=gcd(7378,16762mod7378)=gcd(7378,2006)=gcd(2006,7378mod2006)
=gcd(2006,1360)=gcd(1360,2006mod1360)=gcd(1360,646)=gcd(646,1360mod646)
=gcd(646,68)=gcd(68,646mod68)=gcd(68,34)=gcd(34,68mod34)=gcd(34,0)=34.
2- Find:
a. 13 mod 11=2mod11=2.
b.875mod9=2mod9=2.
c.2594mod48=2mod9=2.
d.217mod21:
Solution:

2. 21mod21=2 22mod21=4, 24mod21=(4x4)mod21=16, 28mod21=(16x16)mod21
=256mod21=4, 216mod21=(4x4)mod21=16,
217mod21=[(21mod21)x(216mod21)]mod21=[2x16]mod21=32mod21=11mod21=11.
3-Using Fermat’s theorem, find 3201mod11?
Solution:
Fermat’s theorem states that:
ap-1 ≡1 mod p , provided that, p is a prime number and a is positive integer not divisible
by p.
3201mod11=[(310mod11)20 x (31mod11)]mod11
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applying Fermat’s theorem
[(1 mod 11)20x (3 mod 11)]mod 11
[1x3] mod 11=3.
4- Perform encryption and decryption using RSA algorithm for the following:
a- p=3 q=11 m=5
Solution
Key generation:
n=p * q=3*11=33.
φ(n)=(p-1)(q-1)=2*10=20.
Select integer “e” where: gcd(φ(n),e)=1 and 1<e< φ(n)
Choose e=3.
Calculate d; where d≡e-1mod φ(n) de≡1 mod φ(n) where: d< φ(n)
d*3≡1 mod 20 d=7.
Public key: KU={3,33}; Private Key: KR={7,33}.
Encryption:
C=Me mod n = 53mod 33=[(51mod33)(52mod33)]mod33=[5*25]mod33 =26.
Decryption:
M=Cd mod n=267mod33=[(261mod33)(262mod33)(264mod33)]mod33
=[26*16*(16*16 mod33)]mod33=[26*16*25]mod33=5.
b- p=13 q=11 M=9.
Key generation:

3. n=p * q=13*11=143.
φ(n)=(p-1)(q-1)=12*10=120.
Select integer “e” where: gcd(φ(n),e)=1 and 1<e< φ(n)
Choose e=7.
Calculate d; where d≡e-1mod φ(n) de≡1 mod φ(n) where: d< φ(n)
d*7≡1 mod 120 721≡1mod120 d=721/7=103.
Public key: KU={7,143}; Private Key: KR={103,143}.
Encryption:
C=Me mod n = 97mod 143=[(91mod143)(92mod143)(94mod143)]mod33
=[9*81*(81*81)mod143]mod143=[9*81*126]mod43=48.
Decryption:
M=Cd mod n=48103mod143
(48mod143)=48.
(482mod143)=(48*48)mod143=16.
(484mod143)=(16*16)mod143=113.
(488mod143)=(113*113)mod143=42.
(4816mod143)=(42*42)mod143=48.
(4832mod48)=(48*48)mod143=16.
(4864mod143)=(16*16mod)143=113.
(4896mod143)=(16*113)mod143=92.
(48100mod143)=(92*113)mod143=100.
(48102mod143)=(100*16)mod143=27.
(48103mod143)=(27*48)mod143=9.
5- In a public key system, an intruder intercepts the cipher text C=25, which is destined
to a user whose public key is {7,133}. What is the plain text message M?
Solution:
We have: KU={e,n}={7,133}
n=133=20305071110130170191=7*19.
p=7, q=19.
φ(n) =(7-1)(19-1)=108.
de ≡ 1 mod φ(n) 7d ≡ 1 mod108.
7d mod 108=1.
Maybe: 7d=108+1? d=109/7=15.57 no
Maybe: 7d=108*2+1 d=217/7=31 OK.
KR={31,133}.
M=Cd mod n =2531 mod 133
251mod133=25.
252mod133=(25*25)mod133=93.
254mod133=(93*93)mod133=4
258mod133=(4*4)mod133=16.
2516mod133=(16*16)mod133=123.
2524mod133=(123*16)mod133=106.
2528mod133=(106*4)mod133=25.

4. 2530mod133=(25*93)mod133=64.
31
M = 25 mod133=(64*25)mod133=4.
M=4.