최소공배수

function smallestCommons(arr) {


  return arr;
}

smallestCommons([1,5]);

function computeSCM(num1, num2) {
  let min = Math.min(num1, num2)
  let max = Math.max(num1, num2)

  for (let i = max; i <= min * max; i+= max) {
    if (i % min === 0) {
      return i; // make sure to find the smallest commmon multiple.
    }
  }
}
// now that we have that we need to find smallest common multiple of an array of numbers or a range.


function smallestCommons(arr) {
  let minNum = Math.min(...arr)
  let maxNum = Math.max(...arr)
  let scm = 1;

  for (let j = minNum; j <= maxNum; j++) {
    scm = computeSCM(scm, j)
  }


  return scm;
}


console.log(smallestCommons([1,5])); will display 60.

function smallestCommons(arr) {
  arr.sort((a, b) => a - b); // comparing two numbers Either way will switch the positions when a is greater than b.


  //arr.sort((a, b => {
    //a > b?-1:1
// }); The conditional (ternary) operator is the only JavaScript operator that takes three operands: a condition followed by a question mark ( ? ), then an expression to execute if the condition is truthy followed by a colon ( : ), and finally the expression to execute if the condition is falsy.
  let [div, num] = arr
  while (div < arr[1]) {
    if (num % div == 0) {
    div++
  } else {
  num += arr[1]; // arr[1] continues being 5 because given the array [1, 5] as an example, arr[1] will always be 5, but the value stored in num increases with each loop from 5 to 10 to 15 to 20 and so on. The first loop tests 5%1, 5%2, 5%3, 5%4 (until it gets a remainder). Second loop tests 10%1, 10%2, 10%3, 10%4. Third loop tests 15%1, 15%2, 15%3, 15%4. And so on until there is no remainder in the loop, which means we have found our solution num. The reason we are not testing divisor 5 is because our dividend is always a multiple of 5 and so we know it will have no remainder when divided by 5 (that's why we can use div < arr[1] instead of div <= arr[1]).
  div = arr[0];
  }
}
  return num
}


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