Stuck need help!The class I'm taking is computer science discrete structures. Problem is attached. please view attachment before answering. Really struggling with this concept. Thank you so much. **Title:** Prove that \(7^n - 1\) is a Multiple of 6 for All \(n \in \mathbb{N}\).**Objective:** Demonstrating that \(7^n - 1\) is divisible by 6 for any natural number \(n\).**Overview:**The statement asserts that for any natural number \(n\), the expression \(7^n - 1\) can be divided by 6 without leaving a remainder. The task involves using mathematical proof techniques to confirm this proposition. **Approach:**To show \(7^n - 1\) is a multiple of 6, it suffices to demonstrate:1. \(7^n - 1\) is divisible by 2.2. \(7^n - 1\) is divisible by 3.Since 6 is the product of 2 and 3, proving divisibility by both numbers will establish the required result.**Potential Mathematical Techniques:**- Mathematical induction- Modular arithmetic**Details of Proof:**1. **Divisibility by 2:** - Consider the parity (odd or even nature) of \(7^n\) and \(7^n - 1\).2. **Divisibility by 3:** - Analyze \(7 \equiv 1 \pmod{3}\), and extend to \(7^n\).The combination of these cases will show \(7^n - 1\) meets the criteria for being a multiple of 6, effectively concluding the proof.

For n = 1 : 71 - 1 is a multiple of 6. therefore, 71 - 1 = 6 * X, (where X belongs to N)…

Answered: Prove that 7" –- 1 is a multiple of 6…

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Prove that 7" –- 1 is a multiple of 6 for all n e N.

Database System Concepts

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ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

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Stuck need help!

The class I'm taking is computer science discrete structures.

Problem is attached. please view attachment before answering.

Really struggling with this concept. Thank you so much.

**Title:** Prove that \(7^n - 1\) is a Multiple of 6 for All \(n \in \mathbb{N}\).

**Objective:** Demonstrating that \(7^n - 1\) is divisible by 6 for any natural number \(n\).

**Overview:**

The statement asserts that for any natural number \(n\), the expression \(7^n - 1\) can be divided by 6 without leaving a remainder. The task involves using mathematical proof techniques to confirm this proposition.

**Approach:**

To show \(7^n - 1\) is a multiple of 6, it suffices to demonstrate:
1. \(7^n - 1\) is divisible by 2.
2. \(7^n - 1\) is divisible by 3.

Since 6 is the product of 2 and 3, proving divisibility by both numbers will establish the required result.

**Potential Mathematical Techniques:**
- Mathematical induction
- Modular arithmetic

**Details of Proof:**

1. **Divisibility by 2:**
- Consider the parity (odd or even nature) of \(7^n\) and \(7^n - 1\).

2. **Divisibility by 3:**
- Analyze \(7 \equiv 1 \pmod{3}\), and extend to \(7^n\).

The combination of these cases will show \(7^n - 1\) meets the criteria for being a multiple of 6, effectively concluding the proof.

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