Trying to solve Project Euler problem 119: The number 512 is interesting because it is equal to the sum of its digits raised to some power: 5 + 1 + 2 = 8, and 8^3 = 512. Another example of a number with this property is 614656 = 28^4. We shall define an to be the nth term of this sequence and insist that a number must contain at least two digits to have a sum. You are given that a2 = 512 and a10 = 614656. Find a30. Question: Is there a more efficient way to find the answer than just checking every number until a30 is found? My Code int currentNum = 0; long value = 0; for (long a = 11; currentNum != 30; a++){ //maybe a++ is inefficient int test = Util.sumDigits(a); if (isPower(a, test)) { currentNum++; value = a; System.out.println(value + ":" + currentNum); } } System.out.println(value); isPower checks if a is a power of test. Util.sumDigits: public static int sumDigits(long n){ int sum = 0; String s = "" + n; while (!s.equals("")){ sum += Integer.parseInt("" + s.charAt(0)); s = s.substring(1); } return sum; } program has been running for about 30 minutes (might be overflow on the long). Output (so far): 81:1 512:2 2401:3 4913:4 5832:5 17576:6 19683:7 234256:8 390625:9 614656:10 1679616:11 17210368:12 34012224:13 52521875:14 60466176:15 205962976:16 612220032:17