Smallest Common Multiple

#javascript #beginners #devops #programming

Now the next problem we're trying to figure out is finding the smallest common multiple of the provided parameters that can be evenly divided by both, as well as by all sequential numbers in the range between these parameters.
For example, if given 1 and 3, find the smallest common multiple of both 1 and 3 that is also evenly divisible by all numbers between 1 and 3. The answer here would be 6.

function smallestCommons(arr) {


  return arr;
}

smallestCommons([1,5]);

Answer:

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]));