You have a few issues with your code. At the moment multiplesOf only accepts 1 argument when it should be two, ie, the numbers array and a single number

Your other issue is that you are not keeping an array of the multiples found, instead, you are setting a variable to the multiples found and is getting overwritten when a new multiple if found (thus leaving you with the very last multiple found in the array). Instead, you want to change your multiples variable to an array. This way you can push every multiple found into this array.

See working example below (read code comments for changes):

function multiplesOf(numbers, number) { // add second argument

  var multiples = []; // change to array (so that we can store multiple numbers - not just one multiple)

  for (var i = 0; i < numbers.length; i++) {

    if (numbers[i] % number === 0) { // divide by the number

      multiples.push(numbers[i]); // add the current multiple found to the multiples array

    }

  }



  return multiples;

}



console.log(multiplesOf([4, 5, 6, 7, 8], 2)); // Output: [4, 6, 8]

Or, if you are happy to use a higher-order function, you can also use .filter() to get your new array. .filter() accepts a function as its first argument which takes an element as its argument. It will keep any of the elements which you return true for in your new array:

const multiplesOf = (numbers, number) => numbers.filter(n => !(n % number));

console.log(multiplesOf([4, 5, 6, 7, 8], 2)); // [4, 6, 8]