Smallest Common Multiple and between

JavaScript
1

Tell us what’s happening:

where did i do wrong

Your code so far


function smallestCommons(arr) {
arr= arr.sort((a,b)=>a-b);
 let a=arr[0];
 let b=arr[1];
 let k=[];
 let d=b;
 let z=0;
 while(z!=arr.length)
 {z=0;
   for(var i=a;i<=b;i++)
 {
  
   if(d%i==0)
   z++;
 }

d++;
 }

  console.log(z);
 return d;
}


console.log(smallestCommons([1,5]));

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Challenge: Smallest Common Multiple

Link to the challenge:

2

Yes u can use more then 1 for

function smallestCommons(arr) {

// find LCM for two numbers
function lowestCommonMultiple(x, y) {
for (let i = x; i <= x * y; i += x) { //generate list of multiples of x
for (let j = y; j <= x * y; j += y) { // generate list of multiples of y
if (i === j) { // find the match
return i;
}
}
}
}
/*
In the for loops the stopping condition is just the multiple of the two numbers,
since that is the last resort case.
I couldn’t see how else to define the end of the loop.
*/

// create array of all numbers from the lower to the higher of input array
function setRange(x, y) {
if (y < x) { // this block corrects for reversed order in input array
const newX = y;
const newY = x;
x = newX;
y = newY;
}
let array = ;
for (let i = x; i <= y; i++) {
array.push(i);
}
return array;
}

// run through array finding LCM of successive pairs
function arrayLowestCommon(array) {
let common = array[0];
for (let i = 0; i < array.length - 1; i++) {
common = lowestCommonMultiple(common, array[i + 1]);
}
return common;
}

const first = arr[0];
const last = arr[1];
const range = setRange(first, last);
const result = arrayLowestCommon(range);
return result;
}