Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 459

For all \(n \geq 2, n^{2}\left(n^{4}-1\right)\) is divisible by (a) 60 (b) 50 (c) 40 (d) 70

Short Answer

Expert verified
The expression \(n^2(n^4 - 1)\) can be factored as \(n^2(n - 2)(n + 2)(n^2 + 4)\). Analyzing the factors, we find that the expression will always be divisible by 60 for all \(n \ge 2\). Thus, the answer is (a) 60.

Step by step solution

01

Factor the expression

First, we need to factor the expression \(n^{2}(n^{4}-1)\). We can rewrite the expression as: \[n^2(n^4-1) = n^2(n^4 - 2^4 + 2^4 - 1)\] Using the difference of squares formula, \((a^2 - b^2) = (a + b)(a - b)\), we can factor further: \[n^2(n^4 - 2^4 + 2^4 - 1) = n^2((n^4 - 2^4) + (2^4 - 1))\] \[= n^2(n^2 - 2^2)(n^2 + 2^2)\] \[= n^2(n - 2)(n + 2)(n^2 + 4)\]
02

Analyze factors

Now, we analyze each factor. - First factor, \(n^2\): Since \(n \ge 2\), \(n^2\) will have at least one factor of 2 and at least one factor of 4. Thus, the first factor is divisible by 4. - Second factor, \((n - 2)\): It has at least one factor of 3 because it is formed by subtracting 2 from a multiple of 3. - Third factor, \((n + 2)\): It has at least one factor of 5 because it is formed by adding 2 to a multiple of 5. - Fourth factor, \((n^2 + 4)\): This factor is a sum of even numbers so it is always even, meaning it has at least one factor of 2.
03

Determine the required multiple

We can now combine the factors of each factor to form the multiple: \[2^2 \times 3 \times 5 \times 2 = 60\] Since 60 contains all the factors we have identified, the expression will always be divisible by 60. Thus, the answer is: (a) 60

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The greatest positive integer, which divides \((\mathrm{n}+2)(\mathrm{n}+3)\) \((n+4)(n+5)(n+6)\) for all \(\forall n \in N\), is (a) 120 (b) 4 (c) 240 (d) 24

For each \(n \in \mathrm{N}, 2 \cdot 4^{2 \mathrm{n}+1}+3^{3 \mathrm{n}+1}\) is divisible by (a) 2 (b) 9 (c) 3 (d) 11

If \(\mathrm{P}(\mathrm{n})\) is a statement such that \(\mathrm{P}(3)\) is true. Assuming \(\mathrm{P}(\mathrm{k})\) is true \(\Rightarrow \mathrm{P}(\mathrm{k}+1)\) is true for all \(\mathrm{k} \geq 3\) then \(\mathrm{P}(\mathrm{n})\) is true (a) for all \(n\) (b) for \(n \geq 3\) (c) for \(\mathrm{n} \geq 4\) (c) none of this

If \(n \in N\), then \(11^{n+2}+12^{2 n+1}\) is divisible by (a) 113 (b) 123 (c) 133 (d) None of these

By principle of mathematical induction, \(\forall \mathrm{n} \in \mathrm{N}\), \(5^{2 \mathrm{n}+1}+3^{\mathrm{n}+2} \cdot 2^{\mathrm{n}-1}\) is divisible by (a) 19 (b) 18 (c) 17 (d) 14
See all solutions

Recommended explanations on English Textbooks

Synthesis Essay

Read Explanation

Rhetorical Analysis Essay

Read Explanation

International English

Read Explanation

Single Paragraph Essay

Read Explanation

Email

Read Explanation

Rhetoric

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free