How to count even odd prime number from 100 to 500?

Question

any hint or solution?

Answer 1

Sorry but your request is not clear..
you would count odd, even or prime numbers from 100 to 500? or all of these?
to determine if a number is odd or even you should use modus operation, but determining primes would be more complex.. please, take a look to my codes to find an hint..

Answer 2

#include <stdio.h>

int main(){
int num=0, i;
for(i=100;i<=500;i++)
(i%2==0)? num++ : num;
printf("Total: %d", num);

return 0;
}

Answer 3

A number x is prime if there's no number n with 1 < n <= sqrt(x) where x%n is 0.
Use the rule above and combine it with a for-loop that tests for every number from 100 to 500 and you have your answer.

Answer 4

With javascript:

var even = [ ];

for(i=500; i>100;--i){

    for(x=i-1;x>=0;--x){

        if(x==1){

            even.push(i);
break;

        }

        else if(i%x == 0 ){

            break;

        }

    }

}

Answer 5

There are no even prime numbers except 2. To count in range use sieve of erathostenes starting from 1, then just count primes in desired range