The Count Factors challenge is just the classic finding the division factors of a number. The most efficient way I found was iterating until the square root of the number and subsequently dividing the number by all the factors we find.

## Solution in Javascript and Complexity **O(sqrt(N))**

```
function solution(N) {
if (N===1) return 1;
let multiplier = 1;
let result = 1;
//can be even more quick if we take out the 2s
for (let i = 2; i <=Math.sqrt(N) ; i++) {
if (N % i === 0) {
multiplier = 1
while (N % i === 0) {
N /= i
multiplier++
}
result *= multiplier
}
}
//if it is a prime number, we add another factor of 2
if (N !== 1) {result *=2}
return result
}
```

Here is the report on Codility. Let me know if there’s a smarter way to solve this!