/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Prove that \((11)^{n}-4^{n}\) is... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 6

Prove that \((11)^{n}-4^{n}\) is divisible by 7 when \(n\) is a natural number.

Short Answer

Expert verified
Since \((11)^n - 4^n \equiv (4)^n - 4^n \equiv 0 \mod 7, \) \((11)^n - 4^n\) is divisible by 7 for any natural number n.

Step by step solution

01

Understand the Problem

We need to prove that \( (11)^n - 4^n \) is divisible by 7 for any natural number n. This means we have to show that the expression \( (11)^n - 4^n \) leaves a remainder of 0 when divided by 7.
02

Reduce the Bases Modulo 7

Simplify the bases 11 and 4 modulo 7. \(11 \equiv 4 \mod 7 \) and \(4 \equiv 4 \mod 7 \). Therefore, the expression can be reduced as follows:\((11)^n - 4^n \equiv (4)^n - 4^n \mod 7.\)
03

Simplify the Expression

Since \( (4)^n - 4^n \equiv 0 \mod 7, \) this means the expression always leaves a remainder of 0 when divided by 7. Hence, \((11)^n - 4^n \) is divisible by 7 for any natural number n.

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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 system of arithmetic for integers, where numbers reset (or 'wrap around') after reaching a certain value鈥攖he modulus. It's like the hours on a clock: after 12 comes 1 again. In this exercise, we use modular arithmetic to simplify numbers before working with larger expressions. For instance, we reduce 11 modulo 7 to simplify our calculations. Since 11 modulo 7 is 4 (because 11-7 is 4), we can easily handle the number in our proof. Understanding this helps us reduce complex expressions and solve problems more efficiently.
natural numbers
Natural numbers are the set of positive integers starting from 1, 2, 3, and so on. They are the basic numbers used for counting and ordering. In our exercise, natural numbers define the values that 'n' can take. Hence, 'n' can be any positive integer. This is important because mathematical properties and proofs often change depending on the set of numbers we are considering.
calculus theory
Although calculus mainly deals with continuous change (differentiation and integration), some foundational ideas help understand discrete mathematics as well. For instance, series and sequences in calculus often provide insights into similar structures in number theory. In our exercise, the series expression \(11^n - 4^n\) might remind us of series in calculus, even though here we deal with discrete natural numbers.
number theory
Number theory is a branch of mathematics dealing with integers and their properties. It includes prime numbers, divisibility rules, and modular arithmetic, among other topics. In our exercise, we employed number theory concepts to prove that \(11^n - 4^n\) is divisible by 7. By reducing our expression using modulo arithmetic and understanding properties of natural numbers, we arrived at a clean and simple proof. Number theory often provides such elegant solutions to what initially seem like complex problems.

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Most popular questions from this chapter

Show that \((2+\sqrt{2})^{1 / 2}\) does not represent a rational number.

Let \(S\) be a nonempty bounded subset of \(\mathbb{R}\). (a) Prove that inf \(S \leq\) sup \(S\). Hint: This is almost obvious; your proof should be short. (b) What can you say about \(S\) if inf \(S=\) sup \(S ?\)

Let \(\alpha\) and \(\beta\) be Dedekind cuts and define the "product": \(\alpha \cdot \beta=\left\\{r_{1} r_{2}:\right.\) \(\left.r_{1} \in \alpha \text { and } r_{2} \in \beta\right\\}\) (a) Calculate some "products" of Dedekind cuts using the Dedekind cuts \(0^{*}, 1^{*}\) and \((-1)^{*}\) (b) Discuss why this definition of "product" is totally unsatisfactory for defining multiplication in \(\mathbb{R}\).

Let \(S\) and \(T\) be nonempty bounded subsets of \(\mathbb{R}\). (a) Prove that if \(S \subseteq T\), then inf \(T \leq\) inf \(S \leq \sup S \leq \sup T\). (b) Prove that \(\sup (S \cup T)=\max \\{\sup S, \text { sup } T\\} .\) Note: In part (b), do not assume \(S \subseteq T.\)

For each set below that is bounded above, list three upper bounds for the set. Otherwise write "NOT BOUNDED ABOVE" Or "NBA." (a) [0,1] (b) (0,1) (c) \\{2,7\\} (d) \(\\{\pi, e\\}\) (e) \(\left\\{\frac{1}{n}: n \in \mathbb{N}\right\\}\) (f) \\{0\\} (g) [0,1]\(\cup[2,3]\) (h) \(\cup_{n=1}^{\infty}[2 n, 2 n+1]\) (i) \(\cap_{n=1}^{\infty}\left[-\frac{1}{n}, 1+\frac{1}{n}\right]\) (j) \(\left\\{1-\frac{1}{3^{n}}: n \in \mathbb{N}\right\\}\) (k) \(\left\\{n+\frac{(-1)^{n}}{n}: n \in \mathbb{N}\right\\}\) (l) \(\\{r \in \mathbb{Q}: r<2\\}\) (m) \(\left\\{r \in \mathbb{Q}: r^{2}<4\right\\}\) (n) \(\left\\{r \in \mathbb{Q}: r^{2}<2\right\\}\) (o) \(\\{x \in \mathbb{R}: x<0\\}\) (p) \(\left\\{1, \frac{\pi}{3}, \pi^{2}, 10\right\\}\) (q) \(\\{0,1,2,4,8,16\\}\) (r) \(\cap_{n=1}^{\infty}\left(1-\frac{1}{n}, 1+\frac{1}{n}\right)\) (s) \(\left\\{\frac{1}{n}: n \in \mathbb{N} \text { and } n \text { is prime }\right\\}\) (t) \(\left\\{x \in \mathbb{R}: x^{3}<8\right\\}\) (u) \(\left\\{x^{2}: x \in \mathbb{R}\right\\}\) (v) \(\left\\{\cos \left(\frac{n \pi}{3}\right): n \in \mathbb{N}\right\\}\) (w) \(\left\\{\sin \left(\frac{n \pi}{3}\right): n \in \mathbb{N}\right\\}\)
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