Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 485

\(10^{\mathrm{n}}+3\left(4^{\mathrm{n}+2}\right)+5\) is divisible by \((\mathrm{n} \in \mathrm{N})\) (a) 7 (b) 5 (c) 9 (d) 17

Short Answer

Expert verified
The expression \(10^n + 3\left(4^{n+2}\right) + 5\) is divisible by 9 (option c).

Step by step solution

01

Test with option (a) 7#

Calculate the expression modulo 7: \[10^n + 3\left(4^{n+2}\right) + 5 \equiv 3^n + 3(4^{n})\pmod{7}\] We will test first for n = 0, n = 1, and n = 2 (we are looking for a pattern): - For n = 0: \(3^0 + 3(4^0) \equiv 1 + 3 \equiv 4 \pmod{7}\) - For n = 1: \(3^1 + 3(4^1) \equiv 3 + 12 \equiv 1 \pmod{7}\) - For n = 2: \(3^2 + 3(4^2) \equiv 9 + 48 \equiv 0 \pmod{7}\) We found a pattern: for n = 2, the expression is divisible by 7. However, further trials would help to confirm if the pattern persists.
02

Test with option (b) 5#

Calculate the expression modulo 5: \[10^n + 3\left(4^{n+2}\right) + 5 \equiv 0 + 2\left(4^n\right) + 0 \equiv 2\left(4^n\right) \pmod{5}\] We will test first for n = 0, n = 1 and n = 2: - For n = 0: \(2(4^0) \equiv 2 \pmod{5}\) - For n = 1: \(2(4^1) \equiv 8 \equiv 3 \pmod{5}\) - For n = 2: \(2(4^2) \equiv 32 \equiv 2 \pmod{5}\) In this case, the expression is not divisible by 5 for any of our tested values of n. Thus, we can eliminate option (b).
03

Test with option (c) 9#

Calculate the expression modulo 9: \[10^n + 3\left(4^{n+2}\right) + 5 \equiv 1^n + 3\left(4^{n}\right) + 5 \equiv 3\left(4^n\right) + 6 \pmod{9}\] We will test first for n = 0, n = 1 and n = 2: - For n = 0: \(3(4^0) + 6 \equiv 3 + 6 \equiv 0 \pmod{9}\) - For n = 1: \(3(4^1) + 6 \equiv 12 + 6 \equiv 0 \pmod{9}\) - For n = 2: \(3(4^2) + 6 \equiv 48 + 6 \equiv 0 \pmod{9}\) In this case, the expression is divisible by 9 for all tested values of n. So, the divisibility by 9 is established.
04

Test with option (d) 17#

Calculate the expression modulo 17: \[10^n + 3\left(4^{n+2}\right) + 5 \equiv (-7)^n + 3\left(4^n\right)\left(4^2\right) + 5 \pmod{17}\] Again, we will test first for n = 0, n = 1 and n = 2: - For n = 0: \((-7)^0 + 3(4^0)(4^2) + 5 \equiv 1 + 48 + 5 \equiv 10 \pmod{17}\) - For n = 1: \((-7)^1 + 3(4^1)(4^2) + 5 \equiv -7 + 192 + 5 \equiv 4 \pmod{17}\) - For n = 2: \((-7)^2 + 3(4^2)(4^2) + 5 \equiv 49 + 3072 + 5 \equiv 8 \pmod{17}\) In this case, the expression is not divisible by 17 for any of our tested values of n. Thus, we can eliminate option (d) too. Based on our tests, the correct answer is option (c) 9. The expression \(10^n + 3\left(4^{n+2}\right) + 5\) is divisible by 9 for the tested values of n.

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

By principle of mathematical induction, \(\forall \mathrm{n} \subset \mathrm{N}, \cos \theta \cos 2 \theta \cos 4 \theta \ldots \cos \left[\left(2^{\mathrm{n}-1}\right) \theta\right]=\) (a) \(\left[\left(\sin 2^{n} \theta\right) /\left(2^{n} \sin \theta\right)\right]\) (b) \(\left[\left(\cos 2^{n} \theta\right) /\left(2^{n} \sin \theta\right)\right]\) (c) \(\left[\left(\sin 2^{\mathrm{n}} \theta\right) /\left(2^{\mathrm{n}-1} \sin \theta\right)\right]\) (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

The product of three consecutive natural numbers is divisible by (a) 6 (b) 5 (c) 7 (d) 4

If \(\mathrm{P}(\mathrm{n}): 3^{\mathrm{n}}<\mathrm{n} !, \mathrm{n} \in \mathrm{N}\), then \(\mathrm{P}(\mathrm{n})\) is true (a) for \(\mathrm{n} \geq 6\) (b) for \(n \geq 7, n \in N\) (c) for \(n \geq 3\) (d) for all \(n\)

For each \(n \in N, 3^{2 n}-1\) is divisible by (a) 8 (b) 16 (c) 32 (d) None of these
See all solutions

Recommended explanations on English Textbooks

Lexis and Semantics Summary

Read Explanation

International English

Read Explanation

Blog

Read Explanation

Research and Composition

Read Explanation

Summary Text

Read Explanation

Language Analysis

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