Faculty of Mathematics University of Waterloo MATH 135
Monday 14 November 2004
MIDTERM EXAM #2 Fall 2005
19:00 20:15
Solutions
[8] 1. Solve the system of congruences x 40 (mod 131) x 7 (mod 18) Solution: Since x 40 (mod 131), then x = 40 + 131y
MATH 135 Assignment 10 Solutions
Problem List
Chapter 9: 5, 23, 40, 44, 45, 48, 58, 63, 100, 159
Fall 2005
Recommended Problems
Exercise 9-5:
Find the sum, difference and product of each of the following pairs of polynomials with coefficients in th
MATH 135 Assignment 9 Solutions
Hand-In Problems
Exercise 8-64:
Express following complex numbers in standard form. the (1 - 3i)8 Solution: |1 - 3i| = 2 and the argument of (1 - 3i) is = 5 , so (1 - 3i) = 2cis 3 By De Moivre's Theorem, we have
MATH 135 Assignment 8 Solutions
Hand-In Problems
Exercise 7-1:
The following is known to be a simple substitution cipher. Break the code.
PCTPG BTHHP TUDAA ANJHT VTHHT DLXCV GDURG CIIDW HXBEA NEIDV XHIGD TBTIW GPEWN DEHLT DS LPHYJ GTSXH AXJHR VJXHT P
MATH 135 Assignment 6 Solutions
Hand-In Problems
Exercise 3-34:
Solve the congruence 5x 7 Solution: Here, gcd(5, 15) = 5, but 5 | 7, so there is no solution by Theorem 3.54. (mod 15).
Fall 2005
Exercise 3-44:
Find the inverse of [23] in Z41 . Solu
MATH 135 Assignment 5 Solutions
Hand-In Problems
Exercise 3-2:
Which of the following integers are congruent modulo 6? -147, -91, -22, -14, -2, 2, 4, 5, 21, 185 Solution: Look at the quotients and remainders on division by 6: x q r -147 -25 3 -91 -16
MATH 135 Assignment 4 Solutions
Hand-In Problems
Exercise 2-9:
If 3p2 = q 2 where p, q Z, show that 3 is a common divisor of p and q. Solution: Since 3 | 3p2 , then 3 | q 2 . By Theorem 2.53, 3 | q or 3 | q, so 3 | q. Let q = 3t, with t Z. Then 3p2
MATH 135 Assignment 2 Solutions
Hand-In Problems
Exercise 1-64: Use the Contrapositive Proof Method to prove that
(S T = ) AND (S T = T ) = S = . Solution: We want to prove the contrapositive of the statement. That is (S = ) = (S T = ) OR (S T =
MATH 135 Assignment 1 Solutions
Hand-In Problems
Exercise 1-12:
Write down the truth tables for each expression. NOT P = (Q R). Solution: P T T T T F F F F Q T T F F T T F F R T F T F T F T F NOT P F F F F T T T T Q R T F F T T F F T NOT P = (Q R)
Faculty of Mathematics University of Waterloo MATH 135
Monday 17 October 2004
MIDTERM EXAM #1 Fall 2005
19:00 20:15
Solutions
[7] 1. (a) Construct truth tables for the two statements P OR (Q OR R) and (NOT P ) AND (NOT Q) = R Solution: For the fir
MATH 135 Assignment 7 Solutions
Hand-In Problems
Exercise 4-28:
A sequence of integers x1 , x2 , x3 , . . . is defined by x1 = 3, x2 = 7, and xk = 5xk-1 - 6xk-2 Prove that xn = 2n + 3n-1 for all n P. Solution: Base Cases When n = 1, x1 = 3 = 21 + 30
MATH 135 Assignment 3 Solutions
Hand-In Problems
Exercise 2-11: Prove that gcd(ad, bd) = |d| gcd(a, b).
Solution: If d = 0, then both sides of the equation are equal to 0, so the result is true.
Fall 2005
Suppose d > 0 and let e = gcd(a, b). Then