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Java Prime Number Method

The question is exactly as follows:

Write a method that determines whether a number is prime.
Then use this method to write an application that determines and displays all the prime numbers less than 10,000.

I have already written a program that finds all prime numbers up to 10,000, but then found a simpler and more efficient one on StackOverflow, and that is this:

package prime;

import java.util.Scanner;

public class Prime
{
public static void main(String args)
{
for(int i = 1; i <= 10000; i++)
{
int factors = 0;
int j = 1;

while(j <= i)
{
if(i % j == 0)
{
factors++;
}
j++;
}
if (factors == 2)
{
System.out.println(i);
}
}
}
}


Since I am very new to Java and am especially not good at methods, this problem is especially difficult for me. I tried making a method, but nothing is being returned, and when I try to return something I get error after error after error.

The help I need is mainly just Pseudo-Code for what I should do to tackle this problem; I'm not asking you for the answer, I'm just asking for a start.

7 Answers
7

package prime;

public class Prime
{
public static void main(String args)
{
for(int i = 1; i <= 10000; i++)
{
if (isPrimeNumber(i))
{
System.out.println(i);
}
}
}


public static boolean isPrimeNumber(int i) {
int factors = 0;
int j = 1;

while(j <= i)
{
if(i % j == 0)
{
factors++;
}
j++;
}
return (factors == 2);
}
}


Write a method that determines whether a number is prime. Then use
this method to write an application that determines and displays all
the prime numbers less than 10,000.

It seems to me that you need a method that returns true if the number you pass it is prime and false otherwise. So, all you really need to do here is to move the portion of the code that deals with determining if a number is prime and, inside that method, instead of printing the number when it is prime, return true, otherwise return false. Then, inside the main method, print the number if that function indicates that the current integer is prime (by returning true) or do nothing otherwise. This is the best I can do to explain the solution without giving you the actual code.

EDIT:Since you asked for pseudocode:

main()
for i in 1..10000 :
if isPrime(i)
print i


isPrime(i):
factors = 0
num = 1

while num is less than or equal to i:
if i mod nums is 0:
increment factors
increment nums
return (factors == 2)


I have found one more solution using java streams:

import java.util.stream.LongStream;

public static boolean isPrime(long num) {
return num > 1 && LongStream.rangeClosed(2, (long)Math.sqrt(num)).noneMatch(div-> num % div== 0);
}


Here goes a test case to prove this works:

// First 168 primes, retrieved from https://en.wikipedia.org/wiki/Prime_number
private static final List<Integer> FIRST_168_PRIMES =
Arrays.asList(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997);

@Test
public void testNumberIsPrime() {
FIRST_168_PRIMES.stream().forEach(integer -> assertTrue(isPrime(integer)));
}

@Test
public void testNumberIsNotPrime() {
List<Integer> nonPrime = new ArrayList<>();
for (int i = 2; i < 1000; i++) {
if (!FIRST_168_PRIMES.contains(i)) {
nonPrime.add(i);
}
}

nonPrime.stream().forEach(integer -> assertFalse(isPrime(integer)));
}


public boolean isPrime(long num)
{
double limit = Math.sqrt(num);
for (long i = 2; i < limit; i++)
if(num%i == 0)
return false;
return true;
}


/**
* List all prime numbers from 101 to 200
* for responding your question, I think it is a good view to have it done
* @ Joannama
*
*/

public class PrimeNum
{
public static void main(String args)
{
for ( int i = 101; i <= 200 ; i += 2)
{
// get rid of even number from 101 to 200
//because even number are never a prime number
boolean f = true;

//assume these numbers are prime numbers
for ( int j = 2; j < i; j++)
{
if ( i % j == 0)
{
f = false;
// if the reminder = 0,
//which means number is not a prime,
//therefore f = false and break it
break;
}
}
if ( ! f)
{
continue;
}
System.out.print("u0020" + i);
}
}
}


As the Brenann posted. Best way to calculate if its a prime, but with only one correction to the limit. You must add 1 to it, or it detects false positives.

public boolean isPrime(long num){
double limit = Math.sqrt(num) + 1;
for (long i = 2; i < limit; i++)
if(num%i == 0)
return false;
return true;
}


static int primeNumber(int n){
int i = 1, factor = 0;
while(i <= n){
if(n % i == 0){
factor++;
}
i++;
}
if (factor == 2){
return 1;
}
return 0;
}


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