A Transition to Advanced Mathematics
A Transition to Advanced Mathematics
8th Edition
ISBN: 9781285463261
Author: Douglas Smith, Maurice Eggen, Richard St. Andre
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter II, Problem 1E

a.

To determine

To write the number as a product of primes and list eight divisors (if composite).

a.

Expert Solution
Check Mark

Answer to Problem 1E

  672=2×2×2×2×2×3×7

Eight divisors of 672 are {2,3,4,6,7,8,14,21} .

Explanation of Solution

Given :

The number is 672 .

672 can be written as a product of primes as follows :

  672=2×2×2×2×2×3×7

Also, the given number is composite.

Hence, the set of any eight divisors of 672 is {2,3,4,6,7,8,14,21} .

b.

To determine

To write the number as a product of primes and list eight divisors (if composite).

b.

Expert Solution
Check Mark

Answer to Problem 1E

  673=1×673 is a prime number.

Explanation of Solution

Given :

The number is 673 .

Since, 673 is a prime number.

Therefore, it can be written as follows :

  673=1×673

c.

To determine

To write the number as a product of primes and list eight divisors (if composite).

c.

Expert Solution
Check Mark

Answer to Problem 1E

  675=3×3×3×5×5

Eight divisors of 675 are {3,9,15,25,27,45,75,135} .

Explanation of Solution

Given :

The number is 675 .

675 can be written as product of primes as follows :

  675=3×3×3×5×5

Also, the given number is composite.

Hence, the set of any eight divisors of 675 is {3,9,15,25,27,45,75,135} .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Let A be a vector space with basis 1, a, b. Which (if any) of the following rules turn A into an algebra? (You may assume that 1 is a unit.) (i) a² = a, b² = ab = ba = 0. (ii) a²=b, b² = ab = ba = 0. (iii) a²=b, b² = b, ab = ba = 0.
No chatgpt pls will upvote
= 1. Show (a) Let G = Z/nZ be a cyclic group, so G = {1, 9, 92,...,g" } with g": that the group algebra KG has a presentation KG = K(X)/(X” — 1). (b) Let A = K[X] be the algebra of polynomials in X. Let V be the A-module with vector space K2 and where the action of X is given by the matrix Compute End(V) in the cases (i) x = p, (ii) xμl. (67) · (c) If M and N are submodules of a module L, prove that there is an isomorphism M/MON (M+N)/N. (The Second Isomorphism Theorem for modules.) You may assume that MON is a submodule of M, M + N is a submodule of L and the First Isomorphism Theorem for modules.
Knowledge Booster
Background pattern image
Advanced Math
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Algebra: Structure And Method, Book 1
Algebra
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:McDougal Littell
Text book image
Holt Mcdougal Larson Pre-algebra: Student Edition...
Algebra
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Text book image
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Text book image
College Algebra (MindTap Course List)
Algebra
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:Cengage Learning
Text book image
PREALGEBRA
Algebra
ISBN:9781938168994
Author:OpenStax
Publisher:OpenStax
Polynomials with Trigonometric Solutions (2 of 3: Substitute & solve); Author: Eddie Woo;https://www.youtube.com/watch?v=EnfhYp4o20w;License: Standard YouTube License, CC-BY
Quick Revision of Polynomials | Tricks to Solve Polynomials in Algebra | Maths Tricks | Letstute; Author: Let'stute;https://www.youtube.com/watch?v=YmDnGcol-gs;License: Standard YouTube License, CC-BY
Introduction to Polynomials; Author: Professor Dave Explains;https://www.youtube.com/watch?v=nPPNgin7W7Y;License: Standard Youtube License