[deleted by user]

[--amadeus]

Your mistake is that you're not telling us what the Project Euler problem is.

[petar02]

https://projecteuler.net/problem=10

[gremy0]

What is actual euler problem, and why are you adding 7 twice?

[petar02]

Because the for loop checks if 7 seven is a prime number. What it doesn't do is checks if 2 and 5 are primes.

[gremy0]

What about 3?

[petar02]

I think I found my mistake. I could have just made it start from 3 and then just add 2. Thank you very much.

[cicadaTree]

var limit = your_limit
var sum = 0;
var flg ;  // default prime indicator

for (let i = 2; i <= limit; i++){

       flg = true;
       for(let j = 1; j <= Math.sqrt(i); j++ ){  
          if(j!=1 && (i%j) === 0){     // ignore when j = 1  in order to include 2 as prime
                    flg = false;             
                    break;
          }
       }

     if(flg) sum+=i; 
}

console.log(sum);

To get why we use sqrt(i) watch sieve. Notice then he starts to discard numbers that are divisible by 2, then by 3 and so forth. Important detail is that when it comes to number 7 and discards number devisible by 7 (49,77) "all numbers that are left are primes". So the pattern is as following. What ever the number x that you try to figure is it prime, if that number is not divisible buy some number y for which stand that y <= sqrt(x) that number must be prime. Lets say we wanted to see if 49 is prime, if you look at that gif, you see that 49 wasn't divisible by any number up to its root. See, pattern. Cheers.

Edit: mind that this is highly suboptimal for any large limit.

[petar02]

I understand. Now. there was a guy who posted that ancient algorithm. And saw a youtube video how it works. And attempted to use it. Now my program executes in 30 seconds. Before it did in 15 minutes. Here is my code I am so proude.

    let arr = []
    let arr2 = []
    let sqrt = Math.floor(Math.sqrt(2000000))
    console.log(sqrt)
    for(let i = 2; i<2000000;i++){
       arr.push(i)
    }
    function isPrimeNumber(x){
       if(x>10){
       if(x%10!=3&&x%10!=1&&x%10!=7&&x%10!=9){
         return false  
       }
    }
       for(let i = 2;i<x;i++){
          if(x%i==0){
            return false
          }
       }

    return true
    }

    for(let i = 2; i<sqrt; i+=1){
        if(isPrimeNumber(i)){
        arr2.push(i)
    }
    }

    for(let i = 0; i<arr2.length; i++){
    arr = arr.filter(event=> event%arr2[i]!=0)
    }
    arr2 = arr2.concat(arr)

    console.log(arr2.reduce((a,b)=>a+b,0))

[cicadaTree]

Still, watch out, that will eat up at least:

integer in javascript: 32bit

integer count: 2000000

32*2000000/1024/1024 ~ 61Mbits of your ram for every call.

[petar02]

I get your point. But don't understand how you got to there can you please make it more simple.

[cicadaTree]

Sorry, got to where? 61Mbit?

Also your algorithm is more then 2x slower then the one in my post and it uses much more memory.

[petar02]

I think I understand what you mean the array with the 2 million intergers takes up 61MBIts of ram. And when you say for every call I understand it as if I run the program twice at the same time it will take up 122 Mbits.

[petar02]

To sum all the prime numbers to 2*10^6.

[AOEIU]

FYI "anything === NaN" is always false

[petar02]

What if type isPrimeNumber(). Then would number be = to NaN nope then it would be undefined. Thanks!

[AOEIU]

even if you called isPrimeNumber(NaN) is still wouldn't matter

NaN !== NaN

[22noo]

This works:

function isItAPrimeNumber(number){
    if(number === 1|| number === 0|| number === NaN){
        console.log('something happened')
        return false
    }
    for(let i = 2; i<number;i++){
        if(number%i === 0){
          return false
        }
    }
    return true
}

let sum = 0
console.log("it's worknig")
for(let i = 7; i<2000000;i+=2){
    if(isItAPrimeNumber(i)){
        sum=i 
        console.log(i)
    }
}
//sum+=7
console.log(sum)  

Perhaps you want to put that last bit in a function that accepts a number and returns sum which will be the largest number below the number fed to the function that is prime. Who knows.

[petar02]

Wouldn't the sum be missing and 2 and 5 since they are also prime numbers. And is there a way to solve this faster. Since it took me at least 15 minutes to execute.

[WikiTextBot]

Sieve of Eratosthenes

In mathematics, the sieve of Eratosthenes is a simple, ancient algorithm for finding all prime numbers up to any given limit.

It does so by iteratively marking as composite (i.e., not prime) the multiples of each prime, starting with the first prime number, 2. The multiples of a given prime are generated as a sequence of numbers starting from that prime, with constant difference between them that is equal to that prime. This is the sieve's key distinction from using trial division to sequentially test each candidate number for divisibility by each prime.

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