Smallest Common Multiple - stuck for days

JavaScript
1

Hey so I’ve been stuck for days in this challenge, seems like my lack of a background in math is taking its toll. As I finally made it to work, it doesn’t pass one of the tests: the one with arguments [1,13] and I can’t figure out why. I’m using firefox and when I test it in debuggers, it work.
Here comes the code:

function smallestCommons(arr) {
  let newArr = arr.sort((a,b) => a-b);
  let divisible = [];
  for (let i = newArr[0]; i <= newArr[1]; i++){
    divisible.push(i);
  }
  console.log(divisible + " divisible arr");
  let mdc = newArr.reduce((a,b) => {
    let resto = 1;
    while (resto !== 0){
    let resto = b % a;
     if (resto > 0){
      b = a;
      a = resto;
    }
  }
    return a;
  });
  let mmc = newArr.reduce((a,b) => {
    return a*b/mdc;
  })
  console.log(mmc + " mmc");
  let result = ((a) => {
    let counter = false;
    for (let i = 2, counter = false; counter === false; i++){    
    let possible = (a*i);
      for (let j = 0, counting = 0; j < divisible.length; j++){
        if (possible % divisible[j] === 0) {
          counting++;
          if (counting === divisible.length){
            return possible;
          }      
        }
      }
    }
  });
  console.log(result(mmc));
  return result(mmc);
}
smallestCommons([1,13]);
2

However inefficient, your code works and passes the tests for me.

I recommend changing your approach.
Right now, you’re multiplying your mmc for every integer in increasing order and then checking if your current number (possible) is divisible by every element in your array (divisible).
Let alone you could make the code more readable:
say, use this:
if (divible.every( x => possible % x === 0) return possible
instead of this:

for (let j = 0, counting = 0; j < divisible.length; j++){
        if (possible % divisible[j] === 0) {
          counting++;
          if (counting === divisible.length){
            return possible;
          }      
        }
      }

But better yet would be to rethink your logic.
Think of a way to combine your mdc and mmc for each element of your array divisible. hThis aricle might help: ttps://stackoverflow.com/questions/147515/least-common-multiple-for-3-or-more-numbers