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 Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

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