fix highly factorable triangular number search to run faster

java/Problem012.java

@@ -4,6 +4,10 @@
  * This program finds the first triangular number with over 500 divisors.
  */
 public class Problem012 implements EulerSolution {
+	public static void main(String[] args) { 
+		Problem012 p = new Problem012();
+		System.out.println(p.run());
+	}
 
 	public String run() {
 		return new Long( findTriangularNumber(500) ).toString();
@@ -11,11 +15,23 @@ public class Problem012 implements EulerSolution {
 
 	/** Find a triangular number with over k divisors. */
 	public static long findTriangularNumber(long k) { 
-		long n = 1;
 		int divisors = 0;
 
-		// We want first divisor with OVER k divisors
+		// Note: the smallest number with 500 factors 
+		// has approx log_2( 500 ) prime factors.
+		// That's 8.9, or rounding down, 8. 
+		// A number with 2^9 prime divisors has 512 factors.
+		//
+		// Next, we know the value of n is proportional to the 
+		// square root of the expression, so we approximate
+		// our starting point by picking a number
+		// that will start at or below the right neighborhood
+		// and work our way up.
+		// 
+		long n = (long)(Math.sqrt(2*3*5*7*11*13*17*19));
 		long tri = 0;
+
+		// We want first number with OVER k divisors
 		while(divisors <= k) { 
 			n++;
 			tri = EulerLib.triangleLong(n);

scratch/012/Triangular.java

@@ -12,17 +12,19 @@ public class Triangular {
 	public static void findTriangularNumber(long k) { 
 		int divisors = 0;
 
-		// Note: the smallest a number with 500 divisors could 
-		// possibly be (strictly speaking, min would be larger, but 
-		// estimate is) 
+		// Note: the smallest number with 500 factors 
+		// has approx log_2( 500 ) prime factors.
+		// That's 8.9, or rounding down, 8. 
+		// A number with 2^9 prime divisors has 512 factors.
 		//
-		// log_2( 500 )
-		//
-		// which is ~8.9
-		//
-		// Start search at 2^9 to speed things up.
+		// Next, we know the value of n is proportional to the 
+		// square root of the expression, so we approximate
+		// our starting point by picking a number
+		// that will start at or below the right neighborhood
+		// and work our way up.
+		// 
+		long n = (long)(Math.sqrt(2*3*5*7*11*13*17*19));
 
-		long n = (long)(Math.pow(2,9));
 
 		// We want first number with OVER k divisors
 		while(divisors <= k) {