判断一个数是否为素数

Question

I just started a java course. The task is to write a code that will output whether a number is prime or not.

This is what I have. It comes up with a lot of errors. I have found a lot of answers but they all say to use lines that we haven't learnt.

public class PrimeNumbers2 {
public static void main(String[] args) {
    int number = 9;
    int count = 2;
    Boolean prime = true;

    if (number % count == 0) {
        prime = true;
    }
    else {

        for (count = count + 1) {

        if (count + 1>= number) {
            prime = false;
        }
        else
        {
            prime =  false;
        }

    }
    if (prime == true)
    {
        System.out.println("Number is prime");
    }
     else
    {
        System.out.println("number is not prime");
    }
}}}

Answer 1

You should think about how you would formulate your problem in natural language. For example: A prime number is a number that is only dividable by itself and 1. or the other way around: For every number in the range of bigger then 1 and smaller then my number, there should be no divider .

To formulate your problem now in simple java code, try to implement your statement.

   for(int count=2; count < number; count++) {
      if(number % count == 0) {
          System.out.println("Number is not a prime!");
          // Since we are finished return and do no more iterations
          return;
      }
   }
   // we iterated all possible numbers and can return successfully
   System.out.println("Number is a prime");

Make also sure to check the edge cases in advance. Eg negative numbers. Please note that is an inefficient solution but the straight forward one.

Answer 2

You have multiple bugs and logic errors in your code. Here it is cleaned up.

int number = 15;
int count = 2;
boolean prime = true;


for (int i = count; i <= number/2; i++ ) {

   if (number % i == 0) {
      prime = false;
  }
}


if (prime == true)
{
    System.out.println("Number is prime");
}
else
{
    System.out.println("number is not prime");
}

Answer 3

If we're talking about a super simple implementation here, you could first check to see if the number is even using the remainder operator, ie, if number % 2 == 0 , then prime = true . If that is not the case, set count = 3 and set up a while loop with the terminating condition while (count < number) . Within the while loop, check to see if the remainder of number and count is zero. If it is, it's a prime number. If not, increment count by one and the loop continues.