commented 1-3

problem1.js

@@ -1,3 +1,9 @@
+//OBJECTIVE:
+//Find the sum of all the multiples
+//of the given values below the limit
+
+//TODO: rewrite using recursion awesomeness
+
 function solve(multiples, limit) {
 	var i=0, sum=0, j;
 	for(;i<limit;i++) {

problem2.js

@@ -1,5 +1,10 @@
+//OBJECTIVE:
+//Find the sum of the even-valued
+//Fibbonacci terms below the given limit
+//This is pretty straighforward, so I'm 
+//not commenting it. :)
 function solve(limit) {
-	var sum=2,i=1,j=fib=2, k=0;
+	var sum=2,i=1,j=fib=2;
 	while(fib<limit) {	
 		fib=i+j;
 		i=j;

problem3.js

@@ -1,14 +1,37 @@
-// factor with recursion
-// time to solve < .5s
+//OBJECTIVE:
+//Find the largest prime factor of the number "num"
 function factor(num, index) {
+	//if the index is null
+	//(because it's our first run on this number)
+	//set it to 2
 	if(!index) index = 2;
+	//the important one here is result=num
+	//i.e, the function returns num by default
+	//if no prime factors are found
 	var lim = Math.ceil(num/2)+1, result=num;
+	//if we're still iterating, basically...
 	if(index < lim) {
+		//if index is a factor at all
 		if(num%index === 0) {
+			//then find its greatest prime factor
+			//recursion!
+			//since the factor function returns the
+			//num passed by default if no prime
+			//factors are found, we'll eventually
+			//reach a point where x === factor(x)
+			//at that point, we've found the 
+			//greatest prime factor, so return it
 			return factor(num/index);
 		}
-		result = factor(num, index+1);
+		//if we're still here, keep going
+		//i.e, set the result to either
+		//the value found in the preceding return 
+		//statement, or to the default value (num)
+		//oh, and increment index so we actually keep
+		//going. :)
+		result = factor(num, ++index);
 	}
+	//yay, we're done! :)
 	return result;
 }