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Chapter 14: Problem 17

6^{n}-1 \text { is divisible by } 5

Short Answer

Expert verified
6^n - 1 is divisible by 5 for any integer n.

Step by step solution

01

Understand the Problem Statement

We need to prove if \(6^n - 1\) is divisible by 5. This means we must demonstrate that the expression \(6^n - 1\) gives a remainder of 0 when divided by 5 for every integer \(n\).
02

Apply Modulus Concept

When checking divisibility, we often use modulus. We need to check if \(6^n \equiv 1 \pmod{5}\). This implies that \(6^n - 1\equiv 0 \pmod{5}\).
03

Simplify Using Modular Arithmetic

Notice that \(6 \equiv 1 \pmod{5}\), since 6 divided by 5 leaves a remainder of 1. Therefore, \(6^n \equiv 1^n \equiv 1 \pmod{5}\) for any positive integer \(n\).
04

Subtract 1 from Both Sides

Since \(6^n \equiv 1 \pmod{5}\), it follows that \(6^n - 1 \equiv 1-1 \equiv 0 \pmod{5}\). Thus, \(6^n - 1\) is divisible by 5.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Modular Arithmetic
Modular arithmetic is a fundamental concept often used to solve problems involving divisibility. It deals with the remainder when one number is divided by another.

When we say that two numbers are congruent modulo a number, it means they leave the same remainder when divided by that number. For example, using the problem from the original exercise, we found that:
  • When 6 is divided by 5, the remainder is 1, which is expressed as \(6 \equiv 1 \pmod{5}\).
  • This congruence helps simplify expressions, like \(6^n \equiv 1^n \equiv 1 \pmod{5}\) for any integer \(n\).
This shows the simplicity of modular arithmetic, where complex exponents can often be reduced to much simpler calculations. Using this method, we can determine divisibility and solve many algebraic problems efficiently.
Integer Exponents
Integer exponents involve raising a number, known as the base, to the power of an integer. In our example, the expression \(6^n\) represents 6 raised to the power of \(n\).

When dealing with exponents in modular arithmetic, a fascinating property emerges:
  • Any number raised to a power respects the modulus of the base. Thus, \(6^n \equiv 1^n \pmod{5}\).
  • The property \(1^n = 1\) makes it evident that no matter the power, the result will always be 1 when the base is congruent to 1.
Understanding integer exponents within the context of modular arithmetic can significantly simplify the problem of determining divisibility. It allows us to predict the outcome of raising numbers to powers and helps in easing complex calculations down to manageable numbers that are easy to handle.
Algebraic Proofs
Algebraic proofs are logical arguments used to establish the truth of mathematical statements. Such proofs often involve a series of logical steps that connect assumptions to a verifiable conclusion.

In the original exercise, we aimed to prove that \(6^n - 1\) is divisible by 5. Here鈥檚 how algebraic proofs come into play:
  • Firstly, by expressing the problem in terms of modular arithmetic: \(6^n \equiv 1 \pmod{5}\).
  • We then demonstrate how the base \(6\) simplifies under modulus to \(1\), making any power of 6 also congruent to 1 modulo 5.
  • Finally, we conclude by showing the expression \(6^n - 1\equiv 0 \pmod{5}\), thus proving divisibility as it leaves a remainder of 0.
Using algebraic proofs provides a structured way to showcase the step-by-step logic behind each calculation, clearly communicating how every piece of the puzzle fits together to establish the truth of the given mathematical statement.

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