Prime test, 2 digit numbers

Question

I want to print all primes 2 digits long. Here is my code:

for(int input = 11; input <= 99; input += 2){ for(int x = 2; x < (int)Math.sqrt(input) + 1; x++){ if(input%x != 0){ System.out.println(input); break; }else{ break; } } } 

The problem is that it prints numbers, such as 35 or 49, which are not prime numbers.

Answer 1

It works. You can see the exit here .

 public class Main { public static void main(String[] args) { for(int input = 11; input <= 99; input += 2){ boolean found = false; for(int x = 2; x < (int)Math.sqrt(input) + 1; x++){ if(input%x == 0){ found = true; break; } } if(!found) { System.out.println(input); } } } } 

Answer 2

Your rudeness test is incorrect. You stop testing the number ( input ) as soon as you find a divisor that input not divisible by.

This is not a prime definition - you need to check that the input number is not divisible by any divisors less than it - in other words, you need to check all the x values ​​before you can declare the number as a prime number.

You can exit the loop that checks input % x != 0 when input is divided by x , but not when it is not divided - you need to continue checking when this condition is true!

Answer 3

I always like questions where the shortest and most obvious answer is to simply burn a hard drive:

 System.out.println("11\n13\n17\n19\n23\n29\n31\n37\n41\n43" + "\n47\n53\n59\n61\n67\n71\n73\n79\n83\n89\n97"); 

Answer 4

To find a prime, you do not need to check every number under it to its square root; you just need to test every other number under it. Here is a demo:

 ArrayList<Integer> primes = new ArrayList<Integer>(); // hold every other prime numbers that we find boolean isPrime; // add first base prime number primes.add(2); for (int x = 3; x < max; x+=2) { // start from 3 and skip multiples of 2 isPrime = true; // prove it not prime for (Integer i : primes) { if (x % i == 0) { isPrime = false; // x is divisible by a prime number... break; // exit loop, we proved it not a prime } } if (isPrime) { if (x >= 10) { System.out.println(x); // print only two digits prime numbers } primes.add(x); // add x to our prime our number list for future checks } } 

** EDIT **

Even more flexible and reusable implementation in the static method (not optimized for listing large primes):

 public static List<Integer> findPrimes(int min, int max) { LinkedList<Integer> primes = new LinkedList<Integer>(); boolean isPrime; double square; // add first base prime number primes.add(2); for (int x = 3; x < max; x+=2) { // start from 3 and skip multiples of 2 isPrime = true; // prove it not prime square = Math.sqrt(x); for (Integer i : primes) { isPrime = x % i != 0; // x is not prime if it is divisible by i if (!isPrime || i > square) { break; } } if (isPrime) { primes.add(x); // add x to our prime numbers } } // remove all numbers below min while (!primes.isEmpty() && primes.getFirst() < min) { primes.removeFirst(); } return primes; } 

Using the method:

 List<Integer> primes = findPrimes(10, 100); System.out.println("Listing " + primes.size() + " prime numbers"); for (Integer prime : primes) { System.out.println(prime); } 

Answer 5

Wikipedia provides a good algorithm for finding primes. It also contains a list of up to 100.

If you are looking for debugging help, I would just skip your code until you see which of your tests are incorrect. This is a simple and simple program that does not take you much time.

Answer 6

The problem is in this block:

 if(input%x != 0) { System.out.println(input); break; } 

It will print 'input' if it is not divided by the current value of 'x', but it can be divided by the following values ​​of 'x'. In this case, "enter" is not simple, but is printed.

The correct code is:

 boolean flag; for(int input = 11; input <= 99; input += 2) { flag = true; for(int x = 2; x < (int)Math.sqrt(input) + 1; x++) { if(input%x == 0) { flag = false; break; } } if(flag == true) System.out.println(input); } 

Answer 7

 for(int input = 11; input <= 99; input += 2){ int found = 0; for(int x = 2; x < (int)Math.sqrt(input) + 1; x++) if(input%x == 0){ found = 1; break; } if(found == 0) System.out.println(input); } 

this should do the job

Answer 8

You print all the odd numbers. You always exit the loop, so x never exceeds 2.

 if(input%x != 0){ System.out.println(input); break; // <<-- break here }else{ break; // <<-- and break here } 

Also, carefully study the definition of a prime number:

• If any of the numbers you are testing is an exact divisor, the number is not prime.
• If all the numbers you are testing are not exact divisors, then the number is prime.

You should exit the loop only if you find a divisor. If you have not found it, you need to continue until you try all the possibilities. And you don’t have to print anything until you finish the cycle and know the result.

Answer 9

Yes, because it prints out all the cases when a non-factor is found. You need ONLY print cases when there are no factors.