# Elementary Number Theory Problems 3.2 Solution (David M. Burton's 7th Edition) - Q1

Ranblog Solution for "Determine whether the integer $701$ is prime by testing all primes $p \leq \sqrt{701}$ as possible divisors. Do the same for the integer $1009$."

## Table of Contents

This is my solution for Chapter 3.2 Q1 in the book *Elementary Number Theory 7th Edition* written by David M. Burton.

## Background

All theorems, corollaries, and definitions listed in the book's order:

**I will only use theorems or facts that are proved before this chapter**. So you will not see that I quote theorems or facts from the later chapters.

## Question

Determine whether the integer $701$ is prime by testing all primes $p \leq \sqrt{701}$ as possible divisors. Do the same for the integer $1009$.

## Solution

For $701$, $p \leq \sqrt{701} \approx 26.48$ means all primes $p \leq 26$.

All primes $\leq 26$ are $2, 3, 5, 7, 11,13, 17, 19,$ and $23$. $701$ is not divisible by any of these. Thus $701$ is prime.

For $1009$, $p \leq \sqrt{1009} \approx 31.76$ means all primes $\leq$ 31.

All primes $\leq 31$ are $2, 3, 5, 7, 11,13, 17, 19, 23, 29$ and $31$. $1009$ is not divisible by any of these. Thus $1009$ is prime.

**Read More:** All My Solutions for This Book

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