## Instructions

**Your Job**

Find the sum of all multiples of n below m

**Keep in Mind**

n and m are natural numbers (positive integers)

m is excluded from the multiples

**Examples**

sumMul(2, 9) ==> 2 + 4 + 6 + 8 = 20

sumMul(3, 13) ==> 3 + 6 + 9 + 12 = 30

sumMul(4, 123) ==> 4 + 8 + 12 + ... = 1860

sumMul(4, -7) ==> "INVALID"

## My solution:

```
function sumMul(n,m){
let r = 0;
for(let i = 1; i*n<m; i++){
r+=i*n
}
return r > 0 ? r : 'INVALID'
}
```

## Explanation

I made an r variable, in which I'll store the sum result.

let r = 0;

Then I used a for loop that will iterate until the result of i*n is smaller than "m".

Inside of this loop I'll change the value of "r" to the sum of "r" plus i*n

for(let i = 1; i*n<m; i++){

r+=i*n

}

At the end if "r" is more than 0, I'll return r, else I'll return 'INVALID'

return r > 0 ? r : 'INVALID'

**What do you think about this solution? ðŸ‘‡ðŸ¤”**

## Top comments (2)

Maybe like this:

This is based on the fact that the average of

`[2, 8]`

is the same as the average of`[2, 4, 6, 8]`

, which means that`(2 + 8) / 2 === (2 + 4 + 6 + 8) / 4`

, where`2 + 4 + 6 + 8`

is the sum you're looking for.Good luck!

Great solution bro, I just tested it and it works ðŸ”¥