# Exam3-Sol - CIS 3362 Test#3 Public Key Encryption Solutons...

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CIS 3362 Test #3: Public Key Encryption Solutons Date: 11/5/2010 1) (8 pts) What is the prime factorization of 1150200? 1150200 = 11502 x 100 = 2 x 5751 x 2 2 x 5 2 = 2 x 3 2 x 639 x 2 2 x 5 2 =2 x 3 4 x 71 x 2 2 x 5 2 . Since 71 is prime, this is our prime factorization: 2 3 x 3 4 x 5 2 x 71 1 . Grading: 2 pts for each prime factor. 2) (8 pts) What is φ(792)? 792 = 2 3 x 3 2 x 11, thus, φ(792) = (2 3 – 2 2 )(3 2 – 3 1 )(11 – 1) = 4 x 6 x 10 = 240 Grading: 1 pt prime factorization, 2 pts for each term in the formula above, 1 pt answer

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3) (12 pts) Using Fermat’s Theorem, determine 171 182 mod 181. Since 181 is prime, Fermat’s Theorem tells us that 171 180 ≡ 1 mod 181. It follows that 171 182 ≡ 171 180 x 171 2 ≡ 1 x (-10) 2 ≡ 100 mod 181. Grading: 6 pts for applying Fermat’s Theorem to 181 and 171. 3 pts for breaking down 171 182 , 2 pts for simplifying that algebra, 1 pt for the answer.
4) (12 pts) Using Euler’s Theorem, determine 17 289 mod 140. 140 = 2

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Exam3-Sol - CIS 3362 Test#3 Public Key Encryption Solutons...

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