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Zhou, HuanXian (0225418)
Zhang, Huan
(0336630)
CS392 HW3
HW #3: Number Theory/Design Principles/Threat Modeling
CS 392/6813: Computer Security
Fall 2010
[100pts] DUE 10/22/2010 at Midnight
Problem 1 [10*3=30pts]
1)
We discussed how to use the Euclidian algorithm to compute the GCD of two
numbers. Use this algorithm to compute the following values. Show all of the steps
involved.
a.
The GCD of 56 and 125.
Euclidian algorithm: If y = ax + b then gcd(x,y) = gcd(x,b)
125 = (a)*56 + b
125 = int(125/56)*56 + 125%56
125 = 2*56 + 13
gcd(56 ,13)
56 = (a)*13 + b
56 = 4*13 + 4
gcd(13,4) = 1
gcd(56, 125) = gcd(56,13) = gcd(13 ,4) = 1
b.
The GCD of 52 and 876.
876 = (a)*52 + b
876 = int(876/52)*52 + 876%52
876 = 16*52 + 44
gcd(52 ,44)
52 = (a)*44 + b
52 = 1*44 + 8
gcd(44, 8)
44 = 5*8 + 4
gcd(8, 4) = 4
gcd(52, 876) = gcd(52 ,44) = gcd(44, 8) = gcd(8,4) = 4
2)
We also discussed the use of the Extended Euclidian algorithm to calculate modular
inverses. Use this algorithm to compute the following values. Show all of the steps
involved.
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View Full Documenta.
957895
1
(mod 1229399450)
Since the gcd of 957895 and 1229399450 is not 1, they are not relatively prime to each
other, thus they don’t have modular inverse.
b.
13
1
(mod 31)
Refer to: http://math.ucdenver.edu/~wcherowi/courses/m5410/exeucalg.html
31 = 2*13+5
p0 = 0
13 = 2*5+3
p1 = 1
5 = 1*3+2
p2 = 0 – 1(2) mod 31 = 2 mod 31 = 29
3 = 1*2+1
p3 = 1 – 29(2) mod 31 = 57 mod 31 = 5
2 = 2*1+0
p4 = 29 – 5(1) mod 31 = 24 mod 31 = 24
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