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Unformatted text preview: 2608977. Now to decrypt, we take the results from above to power d = 1440269 modulo m = 3501697 to get back the numbers we were encoding. This problem has to be done using some programming language, for example Java, since the calculator would not be able to deal with such large numbers. (7) Problem 3 on page 261. (graded) Solution: We have x = 10 + 15 k = 17 + 28 n , for n , k Z . By Euclidean algorithm we get 15(13 7) + 28(7 7) = 7, so x = 10 + 15(91) =1355 mod 420 325 mod 420. So the smallest positive solution is x = 325. (8) Problem 7 on page 261. (graded) Solution: Solving the rst two congruences gives x 38 mod 60. Now solving this equation with x 10 mod 14, gives x 38 mod 420. 1...
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This note was uploaded on 09/05/2011 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 BILLERA
 Algebra

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