## Output prime numbers

There are many algorithms for this task.

Let’s use a nested loop:

```
For each i in the interval {
check if i has a divisor from 1..i
if yes => the value is not a prime
if no => the value is a prime, show it
}
```

The code using a label:

```
let n = 10;
nextPrime:
for (let i = 2; i <= n; i++) { // for each i...
for (let j = 2; j < i; j++) { // look for a divisor..
if (i % j == 0) continue nextPrime; // not a prime, go next i
}
alert( i ); // a prime
}
```

There’s a lot of space to opimize it. For instance, we could look for the divisors from `2`

to square root of `i`

. But anyway, if we want to be really efficient for large intervals, we need change the approach and rely on advanced maths and complex algorithms like Quadratic sieve, General number field sieve etc.

An integer number greater than `1`

is called a prime if it cannot be divided without a remainder by anything except `1`

and itself.

In other words, `n > 1`

is a prime if it can’t be evenly divided by anything except `1`

and `n`

.

For example, `5`

is a prime, because it cannot be divided without a remainder by `2`

, `3`

and `4`

.

**Write the code which outputs prime numbers in the interval from 2 to n.**

For `n = 10`

the result will be `2,3,5,7`

.

P.S. The code should work for any `n`

, not be hard-tuned for any fixed value.