Organized everything, update README

3 files changed, 65 insertions, 0 deletions

diff --git a/12-Highly-Divisible-Triangular-Number/bigfactors2 b/12-Highly-Divisible-Triangular-Number/bigfactors2
new file mode 100755
index 0000000..3e409b1
--- /dev/null
+++ b/12-Highly-Divisible-Triangular-Number/bigfactors2
diff --git a/12-Highly-Divisible-Triangular-Number/bigfactors2.c b/12-Highly-Divisible-Triangular-Number/bigfactors2.c
new file mode 100644
index 0000000..97ba17d
--- /dev/null
+++ b/12-Highly-Divisible-Triangular-Number/bigfactors2.c
@@ -0,0 +1,31 @@
+#include <stdlib.h>
+#include <stdio.h>
+#include <math.h>
+
+int countFactors(int num){
+	int factors = 0;
+    // Check only up until the square root of the number
+ 	int root = (int) ceil(sqrt(num));
+    //printf("%d\n", root);
+    for(int i = 2; i < root; i++){
+        if(num % i == 0)
+            factors+=2;
+    }
+    // Correction for perfect square
+    if(root * root == num)
+        factors -= 1;
+    return factors;
+}
+
+int main(){
+	int i = 1;
+	int k = 1;
+	int j = 0;
+	while(k < 500){
+		j += i;
+		k = countFactors(j);
+		i += 1;
+	}
+	printf("%d has over 500 factors. Neat!\n", j);
+	return 0;
+}
diff --git a/12-Highly-Divisible-Triangular-Number/bigfactors2.py b/12-Highly-Divisible-Triangular-Number/bigfactors2.py
new file mode 100644
index 0000000..9e51502
--- /dev/null
+++ b/12-Highly-Divisible-Triangular-Number/bigfactors2.py
@@ -0,0 +1,34 @@
+import PIL
+import math
+
+# Problem 12 Highly divisible triangular number
+# finds the first number with over 500 factors
+
+
+def countfactors(num):
+    factors = 0
+    # we only need to know about the FIRST HALF of factors,
+    # one factor implies a second
+    root = int(math.ceil(math.sqrt(num)))
+    divs = range(1, root)
+    for d in divs:
+        if(num % d == 0):
+            factors += 2
+
+    # Correction if the number is a perfect square
+    if (root * root == num):
+        factors -= 1
+    return factors
+
+
+####   MAIN #####
+i = 1
+k = 1
+j = 0
+while(k < 500):
+    j += i
+    k = countfactors(j)
+    print(str(j) + " has " + str(k) + " factors")
+    i += 1
+
+print("Ding! Ding! {} has over 500 factors, wow!".format(j))