solutions1 - Spring 2010 MATH 4320: Solutions to Prelim 1...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Spring 2010 MATH 4320: Solutions to Prelim 1 Instructor: Yuri Berest Problem 1. a . Solving the congruence 72 x 36 (mod 376) is equivalent to solving the equation 72 x + 376 y = 36 . Now, using Euclid’s algorithm, we compute (72 , 376) = 8 . Since 8 does not divide 36 , the equation 72 x + 376 y = 36 (and hence the congruence) has no solutions in integers. b . The problem is to find a common solution to the system of three congruences: x 1 (mod 9) , (1) x 3 (mod 7) , (2) x 4 (mod 5) . (3) To do this we use the Chinese Remainder Theorem as follows. First, we solve the first two congruences: it follows from (1) and (2) that x = 1+9 k = 3+7 m for some m, k Z . This gives 9 k - 7 m = 2 ; whence m = k = 1 and x = 10. Thus, by the Chinese Remainder Theorem, a common solution of (1) and (2) is given by x 10 (mod 63) . (4) Now, we find a common solution to the system of congruences (3) and (4). We have x = 4 + 5 s = 10 + 63 t , so that 5 s - 63 t = 6 . Since 5 · ( - 25) + 63 · 2 = 1 , we see that s = - 150, t = - 12 and x = - 746. Thus, a common solution to the system (1)-(3) is x ≡ - 746 (mod 315), or equiva- lently x ∈ {- 746 + 315 k : k Z } . The smallest positive integer in this last set corresponds to
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

solutions1 - Spring 2010 MATH 4320: Solutions to Prelim 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online