function smallestCommons(arr) {
  arr = arr.sort((a, b) => a > b)
  let a = arr[0]
  let b = arr[1]
  let searchingInProgress = true
  let start = 1
  let next = 200
  let numbers = []
  let startIndex = a
  for (let i = 0 ; i < (b - a + 1) ; i++) {
    numbers.push([])
  }
  

  while (searchingInProgress) {
    
    let multiplies = numbers.slice(0)
    //Creating 10x10 2D array of multiplies, each column represents a number in given range
    multiplies = multiplies.map( (number, index) => {
      let subMultiplies = []
      for (let i = start ; i < next ; i++) {
        subMultiplies.push((index + startIndex) * i)
      }
      return subMultiplies
    })
    //Searching through array of multiplies
    multiplies = multiplies.map( (multipliesSequence, index) => {
      //iterate through multiplies in sequence and returns only those which occur in every sequence
       return multipliesSequence.filter( multiply => {
         let alreadyExists = true
         //checks if multiply exists in every column
         for (let i = 0 ; i < multiplies.length ; i++) {
           alreadyExists = alreadyExists && 
                           multiplies[i].indexOf(multiply) !== -1
         }
         return alreadyExists
       })
    })
    console.log(multiplies[0].length)
     if (multiplies[0].length !== 0) console.log(multiplies[0])
    if (multiplies[0].length !== 0) return multiplies[0][0]
    start += next
    next += next
  }
}


smallestCommons([1,5]);