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`hint [3/7]`

Create an array with all the numbers that are missing from the original array to make it easier to check when having to check for even division.

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [4/7]`

You can use modulo `%`

to check if the reminder is 0, which means it is evenly divisible.

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [5/7]`

If you sort the array from greater to lowest then you can check for the first two numbers as it is more likely to the the smallest common multiple than the lower numbers.

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [6/7]`

**Solution ahead!**

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

hint

`hint [1/7]`

Make sure you've tried to hard to solve it yourself before proceeding. :construction:

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [2/7]`

This particular problem can be confusing because most people look for the smallest common multiple of the two number but forget the keyword **range.** This means that if you get `[1,5]`

then you have to check for the smallest common multiple for all these numbers [1,2,3,4,5] that is evenly divisible by all of them.

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [3/7]`

Create an array with all the numbers that are missing from the original array to make it easier to check when having to check for even division.

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [4/7]`

You can use modulo `%`

to check if the reminder is 0, which means it is evenly divisible.

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [5/7]`

If you sort the array from greater to lowest then you can check for the first two numbers as it is more likely to the the smallest common multiple than the lower numbers.

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [6/7]`

**Solution ahead!**

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

```
function smallestCommons(arr) {
return arr;
}
smallestCommons([1,5]);
```

Find the smallest number that is evenly divisible by all numbers in the provided range.

more info:

`bf details`

|`bf links`

|`hint`

`hint [1/7]`

Make sure you've tried to hard to solve it yourself before proceeding. :construction:

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [1/7]`

Make sure you've tried to hard to solve it yourself before proceeding. :construction:

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [2/7]`

This particular problem can be confusing because most people look for the smallest common multiple of the two number but forget the keyword **range.** This means that if you get `[1,5]`

then you have to check for the smallest common multiple for all these numbers [1,2,3,4,5] that is evenly divisible by all of them.

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [3/7]`

Create an array with all the numbers that are missing from the original array to make it easier to check when having to check for even division.

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [4/7]`

You can use modulo `%`

to check if the reminder is 0, which means it is evenly divisible.

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [5/7]`

If you sort the array from greater to lowest then you can check for the first two numbers as it is more likely to the the smallest common multiple than the lower numbers.

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [1/7]`

Make sure you've tried to hard to solve it yourself before proceeding. :construction:

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [2/7]`

This particular problem can be confusing because most people look for the smallest common multiple of the two number but forget the keyword **range.** This means that if you get `[1,5]`

then you have to check for the smallest common multiple for all these numbers [1,2,3,4,5] that is evenly divisible by all of them.

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [3/7]`

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [4/7]`

You can use modulo `%`

to check if the reminder is 0, which means it is evenly divisible.

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [5/7]`

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]

`hint [6/7]`

**Solution ahead!**

type

`hint`

for next hint :pencil: [Contribute at the FCC Wiki]