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I've been having my fun with Project Euler challenges again and I've noticed that my solution for number 12 is one of my slowest at ~593.275 ms per runthrough. This is second to my solution for number 10 at ~1254.593 ms per runthrough. All of my other answers take less than 3 ms to run with most well under 1 ms . My Java solution for Problem 12 : main(): int index = 1; long currTriangleNum = 1; while (numDivisors(currTriangleNum) <= 500) { index++; currTriangleNum += index; } System.out.println(currTriangleNum); numDivisors(): public static int numDivisors(long num) { int numTotal = 0; if (num > 1) if (num % 2 == 0) { for (long i = 1; i * i <= num; i++) if (num % i == 0) numTotal+=2; } else { // halves the time for odd numbers for (long i = 1; i * i <= num; i+=2) if (num % i == 0) numTotal+=2; } else if (num == 0) return 0; else if (num == 1) return 1; else (num < 0) return numDivisors(num *= -1); return numTotal; } . Looking around the solutions forum, some people found that these formulas (n = (p^a)(q^b)(r^c)... & d(n) = (a+1)(b+1)(c+1)...) worked for them, but I personally don't see how it'd be any faster; faster by hand, perhaps, but not in a program. . The basic thought process is as follows: We want to calculate the number of divisors in 48. By looking at the factor tree below, we can conclude that 48 = (2^4)(3^1) [n = (p^a)(q^b)(r^c)...].

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