Assign. 8 Solution (Fall 2008)

Assign. 8 Solution (Fall 2008) - MATH 235 (Basic Algebra)...

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Unformatted text preview: MATH 235 (Basic Algebra) Solutions to Assignment 8 November 15, 2008 Solution of 15. (a) We want to solve the equation 15 x = 11 in Z / 18 Z . But, if x was a solution to the congruence 15 x 11 (mod 18) there exists k Z such that 15 x- 11 = 18 k. Thus 11 = 15 x- 18 k would be divisible by 3, which it isnt. Hence the equation as no solutions. From the point of view of the CRT: we get the solutions in Z / 18 Z by putting together the solutions in Z / 2 Z and in Z / 9 Z , but there are no solutions in Z / 9 Z by the same reasoning. In general we cannot solve ax b (mod n ) if the gcd( a, n ) 6 | b. However, if the gcd( a, n ) | b then we can solve the equation as we will see in the following example. (b) Now we want to solve the equation 15 x = 12 in Z / 63 Z . This equation says that 15 x- 12 is divisible by 63, which is equivalent to saying that 5 x- 4 is divisible by 21, so it suffices to solve the equation 5 x 4 (mod 21). Now 5 is invertible in Z / 21 Z because (5 , 21) = 1. To isolate x , we only need to multiply both sides by the inverse of 5, which is- 4, hence obtaining x - 16 5 (mod 21). So in Z / 63 Z we find the 3 solutions, which are all the x Z / 63 Z with x 5 (mod 21); namely, x = 5 , 26 , 47 . (c) Now the equation is x 2 = 37 in Z / 63 Z . By the CRT, we can find all the solutions modulo 63 by putting together the solutions modulo 7 and modulo 9. Modulo 7, the equation becomes x 2 = 2 which has the solutions x = 3 Z / 7 Z . Modulo 9, it becomes x 2 = 1. A solution of this is given by an integer x such that 9 | x 2- 1 = ( x + 1)( x- 1). Since its impossible that 3 divides both x- 1 and x + 1 (why?), we conclude that the only solutions are x = 1 Z / 9 Z . Now it remains to put these solutions together to get all the solutions in Z / 63 Z . Observe that 4 9- 5 7 = 1 (which we can get by the Division Algorithm). Setting e 1 = 36 and e 2 =- 35, we have e 1 1 (mod 7) , e 1 0 (mod 9) , e 2 0 (mod 7) , e 2 1 (mod 9) , so that all the solutions to our equation are given by 3 e 1 e 2 = 10 , 17 Z / 63 Z (it is easily checked that these are indeed solutions)....
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This note was uploaded on 06/20/2010 for the course LING 200 taught by Professor Noonan during the Winter '09 term at McGill.

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Assign. 8 Solution (Fall 2008) - MATH 235 (Basic Algebra)...

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