Euler in Groovy 7: finding primes

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. What is the 10001st prime number?

It seems a bit cumbersome, but use a sieve of erasthonenes to generate primes until we have collected the 10001st. Basically, iterate up from 3 in steps of 2, check each number against the list of previously encountered primes. If none are divisors, then its a prime, add it to the list.

We save a little time by starting at the first "real" prime (3) and stepping up by 2 from there, because no even number we encounter will be a prime:

primes = [2]
for (i=3; primes.size() < 10001; i+=2) {
        primes_cnt = primes.size()
        found_a_prime = true
        for (j=0; (j < primes_cnt) && (found_a_prime); j++) {
                if (i % primes[j] == 0) {
                        found_a_prime = false
                }
        }
        if (found_a_prime) {
                primes << i
        }
}

println (primes[-1])

which returns 104743