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JavaScript

function smallestCommons(arr) {
var arr0=[]; // source array is formed directly from arguments range
var arr1=[]; // array of multiplies for arr0[i] and numbers in range 1-100
var arr2=[];
arrayOfArrays=[];
var q = [];
var arrayOfMultiplies=[];
var arrayOfCommons=[];
var arrN=arr.sort(
function(a,b){
return a-b;
})
  for (i=arrN[0];i<=arrN[arr.length-1];i++){
  arr0.push(i);} // source array formed 
  
  for (i=1;i<=1000;i++){
  q.push(i);} // numbers to multiply them with source
    //console.log(q);
arr0.forEach(
    function(item, index){
    for (i=0;i<q.length;i++){
  arrayOfMultiplies.push(item*q[i]);
}
arrayOfArrays.push(arrayOfMultiplies);
arrayOfMultiplies=[];
}
)
//no we have arrayOfArrays which stores arrays with numbers for multiplies
   
  
  
    for (i=0;i<arrayOfArrays.length;i++){
  arrayOfArrays[i].forEach(
  function(item, index){
  //console.log(item); //this checks if at this point we properly addressed all multiplies one by one. Yes, we did
  
  if (arrayOfArrays.every(
  (x)=>x.indexOf(item)>-1))
  {
  arrayOfCommons.push(item);}
  }
  )
  }
  
  //at this point we successfully compared each Item with contents of EVERY array due to array.every. Now we need to sort final array.
  
  arrayOfCommons.sort(function(a,b){return a-b;}
  );
  
    
 console.log(arrayOfCommons);
}


smallestCommons([1,13]);