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Exam #3 Review Question Solutions
1) What is the prime factorization of 175?
175 = 5
2
x 7
2) Prove that 2 is a generator mod 13.
2
1
≡ 2 mod 13
2
7
≡ 11 mod 13
2
2
≡ 4 mod 13
2
8
≡ 9 mod 13
2
3
≡ 8 mod 13
2
9
≡ 5 mod 13
2
4
≡ 3 mod 13
2
10
≡ 10 mod 13
2
5
≡ 6 mod 13
2
11
≡ 7 mod 13
2
6
≡ 12 mod 13
2
12
≡ 1 mod 13
This shows that the smallest positive integer for which 2
k
≡ 1 mod 13 is 12, this 2 is a
generator.
3) What is the remainder when 37
129
is divided by 80?
φ(80) = φ((5)(16)) = φ(5) φ(16) = (5 – 1)(16 – 8) = 32. (4 pts)
Thus, 37
32
≡ 1 mod 80.
37
129
≡(37
32
)
4
(37) mod 80 (3 pts)
≡(1)
4
(37) mod 80
≡ 37 mod 80 (1 pt)
4) In an RSA scheme, p = 7, q = 13 and e = 5. What is d?
n = 7x13 = 91
φ(n) = (7 – 1)(13 – 1) = 72. (3 pts)
ed ≡ 1 mod 72
5d ≡ 1 mod 72
Thus, we must find 5
1
mod 72. Use the Extended Euclidean Algorithm:
72 = 14x5 + 2
5 = 2x2 + 1 (3 pts)
5 – 2x2 = 1
5 – 2(72 – 14x5) = 1
5 – 2x72 + 28x5 = 1
29x5 – 2x72 = 1 (3 pts)
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 Fall '08
 Staff
 Information Security

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