3. (i) Find all solutions of the congruence 33x 12 mod 150. (ii) Use part (i) to write all solutions to the equation [33] 150 X = [12] 150 with X € Z/150Z.
3. (i) Find all solutions of the congruence 33x 12 mod 150. (ii) Use part (i) to write all solutions to the equation [33] 150 X = [12] 150 with X € Z/150Z.
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 33EQ
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How do you solve question 3? The second photo provides definitions and examples.
![A set of n integers, containing one representative from each of the n con-
gruence classes in Zn, is called a complete set of residues mod (n). A sensible
choice of such a set can ease calculations considerably. One obvious choice is
provided by the division algorithm (Theorem 1.1): we can divide any integer
a by n to give a = qn+r for some unique r satisfying 0 ≤ r < n; thus each
class [a] € Zn contains a unique r = 0, 1,..., n − 1, so these n integers form
a complete set of residues, called the least non-negative residues mod (n). For
many purposes these are the most convenient residues to use, but sometimes it
is better to replace Theorem 1.1 with Exercise 1.22 of Chapter 1, which gives
a remainder r satisfying -n/2 < r ≤n/2. These remainders are the least ab-
solute residues mod (n), those with least absolute value; when n is odd they
are 0, 11, 12,..., ±(n − 1)/2, and when n is even they are 0, ±1, ±2, ..
±(n − 2)/2, n/2. The following calculations illustrate these complete sets of
residues.
Example 3.3
Let us calculate the least non-negative residue of 28 × 33 mod 35. Using least
absolute residues mod (35), we have 28 = -7 and 33 = -2, so Lemma 3.3
implies that 28 x 33 = (−7) × (-2) = 14. Since 0 ≤ 14 < 35 it follows that 14
is the required least non-negative residue.
Example 3.4
Let us calculate the least absolute residue of 15 x 59 mod 75. We have 15 x 59 =
15 × (-16), and a simple way to evaluate this is to do the multiplication in
several stages, reducing the product mod (75) each time. Thus
15 × (-16) = 15 × (-4) × 4 = (-60) × 4 = 15 × 4 = 60 = -15,
and since -75/2 < -15 ≤ 75/2 the required residue is -15.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F74843223-f587-41ff-9743-6f396250a2ba%2F2c48c682-721c-454a-ad32-aad20891ca13%2Fr5c8qt_processed.png&w=3840&q=75)
Transcribed Image Text:A set of n integers, containing one representative from each of the n con-
gruence classes in Zn, is called a complete set of residues mod (n). A sensible
choice of such a set can ease calculations considerably. One obvious choice is
provided by the division algorithm (Theorem 1.1): we can divide any integer
a by n to give a = qn+r for some unique r satisfying 0 ≤ r < n; thus each
class [a] € Zn contains a unique r = 0, 1,..., n − 1, so these n integers form
a complete set of residues, called the least non-negative residues mod (n). For
many purposes these are the most convenient residues to use, but sometimes it
is better to replace Theorem 1.1 with Exercise 1.22 of Chapter 1, which gives
a remainder r satisfying -n/2 < r ≤n/2. These remainders are the least ab-
solute residues mod (n), those with least absolute value; when n is odd they
are 0, 11, 12,..., ±(n − 1)/2, and when n is even they are 0, ±1, ±2, ..
±(n − 2)/2, n/2. The following calculations illustrate these complete sets of
residues.
Example 3.3
Let us calculate the least non-negative residue of 28 × 33 mod 35. Using least
absolute residues mod (35), we have 28 = -7 and 33 = -2, so Lemma 3.3
implies that 28 x 33 = (−7) × (-2) = 14. Since 0 ≤ 14 < 35 it follows that 14
is the required least non-negative residue.
Example 3.4
Let us calculate the least absolute residue of 15 x 59 mod 75. We have 15 x 59 =
15 × (-16), and a simple way to evaluate this is to do the multiplication in
several stages, reducing the product mod (75) each time. Thus
15 × (-16) = 15 × (-4) × 4 = (-60) × 4 = 15 × 4 = 60 = -15,
and since -75/2 < -15 ≤ 75/2 the required residue is -15.
![1. (i) Find a complete set of residues (i.e., congruence class representatives) mod 13 con-
sisting of odd integers.
(ii) Suppose that gcd (m, n) = 1. Show that there exists a complete system of residues mod n
all of which are divisible by m.
In the next two problems, please do the calculations by hand and show your work.
2. (i) Find all solutions of the congruence
33x 12 mod 51.
(ii) Use part (i) to write all solutions to the equation
[33]51 X = [12]51
with X € Z/51Z.
3. (i) Find all solutions of the congruence
33x 12 mod 150.
(ii) Use part (i) to write all solutions to the equation
[33]150 X = [12]150
with X € Z/150Z.
4. Let
f(x) = x + a₁xm-1+. ·+am-1x + am,
aj € Z,
be a polynomial with integer coefficients and m≥ 1.
(i) Show that if a and b are integers with a = b mod n, then f(a) = f(b) mod n.
(ii) Suppose that for some integer a, f(a) = p is a prime. Show that, if f(b) = q is also a
prime for some integer b with b = a mod p, then p = q.
(iii) As in part (ii), suppose that for some integer a, f(a) = p is a prime. Show that there
must be an integer b with f(b) not a prime. (Hence there are no 'magic' polynomials, giving
only prime values. ) Hint: consider the polynomial f(x) -p and use (ii).
1This means all integers x that satisfy the congruence.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F74843223-f587-41ff-9743-6f396250a2ba%2F2c48c682-721c-454a-ad32-aad20891ca13%2Fxdcbayk_processed.png&w=3840&q=75)
Transcribed Image Text:1. (i) Find a complete set of residues (i.e., congruence class representatives) mod 13 con-
sisting of odd integers.
(ii) Suppose that gcd (m, n) = 1. Show that there exists a complete system of residues mod n
all of which are divisible by m.
In the next two problems, please do the calculations by hand and show your work.
2. (i) Find all solutions of the congruence
33x 12 mod 51.
(ii) Use part (i) to write all solutions to the equation
[33]51 X = [12]51
with X € Z/51Z.
3. (i) Find all solutions of the congruence
33x 12 mod 150.
(ii) Use part (i) to write all solutions to the equation
[33]150 X = [12]150
with X € Z/150Z.
4. Let
f(x) = x + a₁xm-1+. ·+am-1x + am,
aj € Z,
be a polynomial with integer coefficients and m≥ 1.
(i) Show that if a and b are integers with a = b mod n, then f(a) = f(b) mod n.
(ii) Suppose that for some integer a, f(a) = p is a prime. Show that, if f(b) = q is also a
prime for some integer b with b = a mod p, then p = q.
(iii) As in part (ii), suppose that for some integer a, f(a) = p is a prime. Show that there
must be an integer b with f(b) not a prime. (Hence there are no 'magic' polynomials, giving
only prime values. ) Hint: consider the polynomial f(x) -p and use (ii).
1This means all integers x that satisfy the congruence.
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