freeCodeCamp Challenge Guide: Sum All Primes

Guide
47 
function sumPrimes(num) {
  var sumPrimez = 0;
  var count = 0;
  var arr = [0];
  for(var i=1;i<=num;i++){
    for (var j=1;j<=i;j++){
      if((i%j)==0){
        count+=1;
      }      
    }
  arr.push(count);
  count = 0;
}
for(var i=0;i<arr.length;i++)
{
    if (arr[i]==2){
        sumPrimez+=i;
    }
}
return sumPrimez;
}

sumPrimes(977);

Simple but not so efficient. How do you guys come up with these way cooler algorithms ? :tired_face:

48

function sumPrimes(num) {
var i=1,sum=0;
while(num>=i){
if(isPrime(i))sum+=i;
i++;
}

return sum;
}
function isPrime(num) {
for(var i = 2; i < num; i++)
if(num % i === 0) return false;
return num !== 1;
}
sumPrimes(10);

here’s my code

49 
function sumPrimes(num) {
  var prim=[],primtst;
 for(var i=2 ;i<=num;i++){
  primtst=0;
   for(var j=1;j<=i;j++){
   if(i%j==0){
    primtst++;
   }
   }
  if(primtst==2){
    prim.push(i);
  }
 

}
return prim.reduce(function(a,b){return a+=b;});
}

sumPrimes(10);
50

Hello everyone,

'Cause I didn’t saw something exactly like this, I think I’ll add my solution in the pool:

    function sumPrimes(num) {
      
      var temp = [],
          primes = [],
          sumUp;
      
      function zPrime(num) {
        for (var x = 2; x < num; x++) {
          if(num < 2) return false;
          if(num % x === 0) return false;
        } return true;
      }  
      for (var y = 2, z = 0; y <= num, z < num-1; y++, z++){
        if(y <= num) temp.push(y);
        if(zPrime(temp[z]) === true) primes.push(temp[z]);
      }
      sumUp = primes.reduce(function(a,b){
        return a + b;
      });
      return sumUp;
    }
    //There is no spoon!
    sumPrimes(1000);
51

stackoverflow pointed me to two different methods

function sumPrimes(num) {
  var PrimeArr=[2];
  if (num <= 1){
    return false;
  }else if (num === 2){
    PrimeArr.push(2);
  } else { 
    for (var i=3;i<=num;i++){
    for (var j=2;j<i;j++){
      if (i%j===0){break;}
      if (j===i-1){
      PrimeArr.push(i);}
    }
    }
      }
 
  return PrimeArr.reduce(function (sum,value){
    return sum +=value;
  });
}

second one is sqrt method

function sumPrimes(num) {
  var PrimeArr=[2,3];
  if (num <= 1){
    return false;
  }else if (num === 2){
    PrimeArr.push(2);
  } else {
    for (var i=2;i<=num;i++){
    var sq=Math.floor(Math.sqrt(i));
    for (var j = 2; j <= sq; j++){
        if (i % j === 0){break;}
        if (j===sq){
        PrimeArr.push(i);}
    }
    }
      }

  return PrimeArr.reduce(function (sum,value){
    return sum +=value;
  });
}
1 Like
52 
function sumPrimes(num) {
  
  //[STEP 1]An array to store all numbers from 0 - 'num'
  var arr=[];
  
  //[STEP 2]A variable to store the sum of all primes numbers
  var total = 0;

  /*[STEP 3] Loops 'num' number of times and 'pushes' all
    numbers from 0 to 'num' into the array 'arr' */
  for(var j = 1; j <= num; j++){
   arr.push(j); 
  }
  
 //[STEP 4.1] loop through the array 'arr' to test each number for prime
 for(var i = 0; i < arr.length; i++) {
   /*[STEP 4.2] pass each number from the 
     array 'arr' into the 'isPrime' function (defined below)
     to check if it is a prime number.
     (Go to isPrime function below for next step) */
   if(isPrime(arr[i]) == true){
 
     //[STEP 6] If 'true' the number is added to total
     total+= arr[i];
   }
  }
  
  /*[Step 7] After loop is complete, return 'total' 
    which stores the sum of all the primes number until 'num' */
  return total;
    }

//[STEP 5] returns true or false if the number is prime or not.
function isPrime(value) {
    for(var i = 2; i < value; i++) {
        if(value % i === 0) {
            return false;
        }
    }
    return value > 1;
}

sumPrimes(10);
53

What about this solution?

function isPrime(num) {
  for (let i = 2; i < num; i++) {
    if (num % i === 0) {
      return false;
    }
  }
  return true;
}

function sumPrimes(num) {
  var sum = 0;
  for (let i = 2; i <= num; i++) {
    if (isPrime(i)) {
      sum += i;
    }
  }
  return sum;
}
8 Likes
56 
function sumPrimes(num) {
  let primeArr = [];
  nextPrime:
  for (let i = 2; i <= num; i++) {
    for (let j = 2; j < i; j++) {
      if (i % j == 0) continue nextPrime;
    }
    primeArr.push(i);
  }
  return primeArr.reduce((acc, curr) => acc + curr);
}

sumPrimes(10);
2 Likes