# freeCodeCamp Bonfire: Smallest Common Multiple

Oh man, this one was a doozy. Here’s the assignment:

Find 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.

The range will be an array of two numbers that will not necessarily be in numerical order.

e.g. for 1 and 3 - find the smallest common multiple of both 1 and 3 that is evenly divisible by all numbers between 1 and 3.

Sounds easy, right? It pretty difficult for me to wrap my head around, but I actually got this one **almost** on my first try. See my solution below with explanations of each step in the comments.

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function smallestCommons(arr) {
// Define two arrays to get the full sequence for both numbers in the array
var sequence1 = [];
var sequence2 = [];
// I'm going to use this variable to tell this function to keep running the loop
var num = false;
// I'll use this variable as a counter to check each potential multiple
var counter = 1;
// These two for loops are building my two sequences
for (var i = 1; i <= arr[0]; i++) {
sequence1.push(i);
}
for (var j = 1; j <= arr[1]; j++) {
sequence2.push(j);
}
// Here's the meat of the operation, running a while loop until the answer is found
while (num === false) {
// Check to see if the current number is a multiple of both starting values
if (counter % arr[0] === 0 && counter % arr[1] === 0) {
// Divide the current number by every number in both sequences,
// checking to see if they're evenly divisible. If they are, all
// the values in these two new arrays will be 0
var check1 = sequence1.map(function(x) {
return counter % x;
});
var check2 = sequence2.map(function(x) {
return counter % x;
});
// Check to see if all values in both new arrays are 0. If they are,
// set num = true to stop running the loop.
if (check1.every(elem => elem === 0) && check2.every(elem => elem === 0)) {
num = true;
}
}
// Increase the current number only if loop is still set to run
if (!num) {
counter++;
}
}
return counter;
}

Like the last few bonfires I’ve worked on, I get done this one, I look at the code and I think to myself: “There’s got to be a simpler way to do this.” I’m going to make it a habit to look for solutions other people have found after I finish eash bonfire, and I hope these posts serve that purpose for others too.