I've changed the while into a for like you asked, and also made the algorithm twice as fast by skipping the evaluation of even numbers which are all non-prime. Actually it's more than twice as fast with several other optimizations I made.

Code:
#include <stdio.h>

int main(void)
{
    int i=0,j=0,l=0;

    printf("1 is a prime number\n");
    printf("2 is a prime number\n");

    for (i=3; i < 101; i=i+2)
    {
        l = 0;

        // skip 1, don't go as high as i
        for (j=2; j<i; j++)
        {
            if (i % j == 0)
            {
                l++;
                // we already know it isn't prime
                break; 
            }
        }

        if (l == 0)
        {
            printf("%d is a prime number\n", i);
        }
    }

    return 0;
}
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