Here’s my code. (I’m sure there’s a better way to do this, but I want to muscle my way through on my own before looking at other solutions.)

```
function smallestCommons(arr) {
var lesser, greater;
if (arr[0] < arr[1]) {
lesser = arr[0];
greater = arr[1];
} else {
lesser = arr[1];
greater = arr[0];
}
var factorsArray = [];
var multiplesOfGreater = [];
for (var i = lesser; i <= greater; i++) {
factorsArray.push(i); //put in the integers between the two values
}
var limit = factorsArray.reduce(function(a, b) {
return a * b;
}); //the SCM is never going to be more than the product of all the numbers.
for (var j = 1; j <= limit / greater; j++) { //we're going to start making multiples of greater, starting with itself x 1. We will go until we hit the limit (i.e. until j is a number that, multiplied with greater, equals limit).
var currentMultiple = greater * j;
for (var k = 0; k < factorsArray.length; k++) { //now we'll ask whether this multiple is divisible by each of the integers in factorsArray in turn.
if (currentMultiple % factorsArray[k] !== 0) {
break; //as soon as it's not divisible by one of the integers, our work here is done and let's consider a different multiple. Break this for loop (k) and do the next iteration of the outer one (j)
} else {
//if it is divisible, go on to check the next integer. Repeat this for loop with a different k.
}
return currentMultiple; //if it gets this far, it's passed every integer; this is the answer
}
}
}
```

My premise, with those nested for loops, is “if we can get through all of `factorsArray`

, proving that the multiple of `greater`

in question is divisible by each, then it’s our answer.” But if I put the `return`

line inside the second for loop, it never gets to it because it ends the final iteration with `continue;`

and peters out. If I put it inside the first, outer for loop, it ends the whole thing after the very first `break;`

and returns nothing. How can I make this work?