function smallestCommons(arr) {
let multiple = [];
for (let i = 0; i < arr.length; i++) {
if (arr.range(i,i + 1) % 2 == 0) {
multiple % arr.range(i,i + 1) === 0;
}
}
return multiple;
}
smallestCommons([1,5]);

My head is all over the place with this one. I thought I could use the range method in this way but I am doing a lot of things wrong. Can someone help me out?

I would ignore fancy syntax. You only need to express a clear, detailed plan with if and for. Fancy syntax cannot make a plan for you if you donâ€™t have one.

Iâ€™d suggest you refactor your code by setting the conditions clear. After setting the condition, donâ€™t forget the empty multiple array. Check string methods i.e Push()
Good luck buddy!

function smallestCommons(arr) {
for (let i = 0; i < arr.length; i++){
// if the two numbers in arr can be evenly divided
if (i % 2 == 0 && (i + 1) % 2 == 0)
// if all the numbers between them can also be evenly divided
// return the first smallest common multiple all these numbers have
}
return arr;
}
smallestCommons([1,5]);

Is this a better straightforward plan? I checked if the numbers in the arr can be evenly divided, but I am having a hard time thinking of a way to check all the numbers in between as well

function smallestCommons(arr) {
// Check all the multiples of numbers 1-5
// a multiple of a number is basically adding up by that number over and over
// for example 1:[1,2,3,4,5,6,7,8,9...]
// 2: [2,4,6,8,10...]
// 3: [3,6,9,12,15,18,21...]
// 4: [4,8,12,16,20...]
// 5: [5,10,15,20,25,30...]
// if you continued going up, eventually you would realize that 60 is the smallest common multiple for numbers 1-5
return arr;
}
smallestCommons([1,5]);

I really appreciate how blunt and honest you are. I didnâ€™t write any code at all and wrote out the process, is this better?

Because you would see that 60 is the first number that 1,2,3,4, and 5 all have in common. Like 1-3 all have 6 in commonâ€¦but you canâ€™t say thatâ€™s the smallest common multiple because 4 and 5 donâ€™t have 6

I can see this by looking at the pattern I have made for each number. Or do you mean like how can you see this in terms of code? Maybe by creating a loop with instructions for each number that tells the number to keep adding by itâ€™s own value until all numbers find a number they have in common

The smallest common multiple is going to be divisible by all of those numbers (2,3,4,5). We can guess the right number will have a remainder of 0 when divided by (2,3,4,5).

We could only look at the largest numberâ€¦5. Then once 5 shows us a number that is also divisible by 4,3,2â€¦then we know that is the lowest common multiple. Instead of looking through each numbers sequence