HOMEWORK 8
Math 4132
#2, p.185
Solution
a) There are 7 solutions.
x 0, y 4 (mod 7)
x 1, y 1 (mod 7)
x 2, y 5 (mod 7)
x 3, y 2 (mod 7)
x 4, y 6 (mod 7)
x 5, y 3 (mod 7)
x 6, y 0 (mod 7)
b) No solution.
#4, p.222
Solution
By Wilsons Theorem, 30! 1 30 (mod 3
Some practice problems for midterm 2
Kiumars Kaveh
November 15, 2011
Problem: What is the remainder of 62000 when divided by 11?
Solution: This is a long-winded way of asking for the value of 62000 mod 11.
Since 11 is prime and 6 is not a multiple of 11,
Some answers for Homework 12
Math 406
Here are solutions to some of the trickier problems :
Section 7.1, # 22: Show that if the equation (n) = k where k is a positive integer has
exactly one solution n, then 36|n.
Proof. We proceed in fours steps; rst we
Some practice problems for midterm 2
Kiumars Kaveh
November 15, 2011
Problem: What is the remainder of 62000 when divided by 11?
Solution: This is a long-winded way of asking for the value of 62000 mod 11.
Since 11 is prime and 6 is not a multiple of 11,
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THEORY OF NUMBERS
HOMEWORK 8 SOLUTIONS
Computations
(1) Find a primitive root mod n for each of the following n:
(a) 232
(b) 192
(c) 13k for any k
(d) 37k for any k
(e) 38
(f) 54
Solution:
(a) Since (23) = 22, there are four possible numbers for the order
MTH 382
Number Theory Practice Problems for the Final
Spring 2003
(1) Find the quotient and remainder in the Division Algorithm, (a) with divisor 16 and dividend 95, (b) with divisor 16 and dividend -95, (c) with divisor [6.2] and dividend 95, (d) with di
MT 430EXAM 1SOLUTIONS
NAME:
DATE:
1. The following exercise pertains to the polynomial f (x) = x3 + 3x2 + 5x + 5.
a. Find a complete list of incongruent solutions to f (x) 0 (mod 7). [8 pts]
We just compute f (a) for 0 a 6 and nd
f (0) = 5 0
mod 7
f (1) =
MTH 382
Number Theory Practice Problems for the Final
Spring 2003
(1) Find the quotient and remainder in the Division Algorithm, (a) with divisor 16 and dividend 95, (b) with divisor 16 and dividend -95, (c) with divisor [6.2] and dividend 95, (d) with di
Math 116 Number Theory
with Travis Kelm
Homework #5
Congruences
Spring 2007
CSU - Fresno
Due 1 March 2007
Solutions
In your solutions you must explain what you are doing using complete sentences.
Section 4.3 - The Chinese Remainder Theorem
Exercise 4abc:
HOMEWORK ASSIGNMENT #4 SOLUTIONS
(1) (a) Find all simultaneous solutions to the system x 7 mod 20, x 2 mod 3.
(b) Find all simultaneous solutions to the system x 3 mod 7, x 2 mod 8, x
1 mod 5.
Solution.
(a) The CRT tells us there is a unique solution mod
University of Toronto Mississauga
Departement of Mathematics
Instructor: K. Shaw
TA: A. Mousavidehshikh
Introduction to Number Theory - MAT315H5
Problem set 6 - Due February 28, 2012
Exercise 1. Show that if n is odd, then n2 1 (mod 8). Is the number 8293
MAT 301 Problem Set 1: Solutions
Posted: January 6, 2012
Due: January 16, 2012
Worth: 100 points
Problem 1: Number Theory Review (60 points)
1. (15 points) Compute the multiplicative inverse of a (mod n) for the following values of a
and n:
a = 5, n = 17
MAT 311: Number Theory Spring 2006
Solutions to HW5
1. (Davenport, pp.217, ex. 2.05) We would like to nd the remainder when x := (10273 + 55)37 is divided by 111 = 3 37. To do this, we rst nd the remainders mod 3 and mod 37; for then, those remainders wil
Math 453, Section X13
Final Exam Solutions
Spring 2011
Problem 1
True/false questions. For each of the following statements, determine if it is true or false, and provide
a brief justication for your claim. Credit on these questions is based on your justi
MATH 115A SOLUTION SET III JANUARY 27, 2005 (1) Use Fermat's Little Theorem to verify that 17 divides 11104 + 1. Solution: Working modulo 17, we have that 11104 (112 )52 252
5 10
(mod 17) (mod 17) (mod 17)
(mod 17)
10
(2 ) 22 (-2) 4 44
(mod 17)
16 (mod
SOLUTIONS TO HOMEWORK 5:
7.2 (4) Establish the following:
(a) If n is an odd integer then (2n) = (n).
Since is multiplicative and gcd(2, n) = 1, we know (2n) = (2)(n).
But (2) = 1, so (2n) = (n).
(b) If n is even then (2n) = 2(n).
If n is even then n = 2k
115 Homework 9
Due Friday December 3
Question 1 (Rosen 7.2.8,9,10,11) Which positive integers have exactly (i) two
positive divisors, (ii) three positive divisors and (iii) four positive divisors? What
is the product of positive divisors for any
? (Why?)
Koushik Pal
Math 406, Midterm II
April 12, 2012
Problem 1.
(a) Find the remainder when 25! is divided by 29.
(b) Show that 42|(n7 n) for all positive integers n.
(15 + 15 = 30 points)
Solution. (a) By Wilsons Theorem, 28! 1 (mod 29). But
28! = 28 27 26 25
PMTH338 NUMBER THEORY
The tutorial questions contained in this booklet have mostly been selected from
Elementary Number Theory and its Applications by Kenneth H. Rosen, 3rd ed.
Addison Wesley. References to relevant sections from this text are included wi
Proposition 2.1.5. Let a and b be integers, not both zero. Then any common divisor of a and
b is a divisor of gcd(a, b).
Proof. The cast of characters in this proof:
Integers a and b such that a2 + b2 > 0 .
By Proposition 1.3.8 there exists a greatest c
Math 25: Solutions to Homework # 5
(6.2 # 8) Show that if p is prime and 2p 1 is composite, then 2p 1 is a pseudoprime to
the base 2.
Let m = 2p 1. Since p is prime, 2p 2 (mod p), so p | 2p 2, hence 2p 2 = kp for
p
some integer k . Then 2m1 = 22 2 = 2kp .