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Problem Explanation:
The explanation for this problem is very simple. You will generate a list of prime numbers up to the number you are given as a parameter. Then you need to add them all up and return that value. The tricky part is on generating the list of prime numbers. I suggest you find a code or a good math algorithm that you can turn into code.
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Hint: 1
Generate a list of all the numbers up to and including the one you got as a parameter. This will be needed to determine which numbers are prime or not.
try to solve the problem now
Hint: 2
Check this link if you prefer to find a solution for finding primes, or try learning and implementing your own Sieve of Eratosthenes
try to solve the problem now
Hint: 3
This problem is hard if you have to create your own code to check for primes, so don’t feel bad if you had to use someone’s code for that bit. Either way, you are most likely using array, so once you generate an array of primes, then just add them all up and return the number you get.
try to solve the problem now
Spoiler Alert!
Solution ahead!
Basic Code Solution:
function sumPrimes(num) {
var res = 0;
// Function to get the primes up to max in an array
function getPrimes(max) {
var sieve = [];
var i;
var j;
var primes = [];
for (i = 2; i <= max; ++i) {
if (!sieve[i]) {
// i has not been marked  it is prime
primes.push(i);
for (j = i << 1; j <= max; j += i) {
sieve[j] = true;
}
}
}
return primes;
}
// Add the primes
var primes = getPrimes(num);
for (var p = 0; p < primes.length; p++) {
res += primes[p];
}
return res;
}
// test here
sumPrimes(10);
Code Explanation:
 Create a function that generates the numbers from 1 to num and check if they are prime along the way.
 Declare the variables that will be needed.
 Start with 2, if it has not been marked and added to the sieve array then it is a prime and we add it to the prime array.
 Add the others to the sieve array.
 Return the primes
 Loop through the returned array and add all the elements to then return the final value.
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Intermediate Code Solution:
function sumPrimes(num) {
// function to check if the number presented is prime
function isPrime(number){
for (i = 2; i <= number; i++){
if(number % i === 0 && number!= i){
// return true if it is divisible by any number that is not itself.
return false;
}
}
// if it passes the for loops conditions it is a prime
return true;
}
// 1 is not a prime, so return nothing, also stops the recursive calls.
if (num === 1){
return 0;
}
// Check if your number is not prime
if(isPrime(num) === false){
// for non primes check the next number down from your maximum number, do not add anything to your answer
return sumPrimes(num  1);
}
// Check if your number is prime
if(isPrime(num) === true){
// for primes add that number to the next number in the sequence through a recursive call to our sumPrimes function.
return num + sumPrimes(num  1);
}
}
// test here
sumPrimes(10);
Code Explanation:
 The function
isPrime
checks if a particular number is prime or not.  If
num
is 1, return 0 since 1 is not a prime number.  If num is not prime, check next number down from maximum number.
 If num is prime, add it to next number in the sequence through recursion to
sumPrimes
function.
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Advanced Code Solution:
function sumPrimes(num) {
// step 1
let arr = Array.from({length: num+1}, (v, k) => k).slice(2);
// step 2
let onlyPrimes = arr.filter( (n) => {
let m = n1;
while (m > 1 && m >= Math.sqrt(n)) {
if ((n % m) === 0)
return false;
m;
}
return true;
});
// step 3
return onlyPrimes.reduce((a,b) => a+b);
}
// test here
sumPrimes(977);
Code Explanation:

Step 1: Use
Array.from()
to generate a sequence of numbers up to and includingnum
. Combine with.slice()
to slice off first two indices[0, 1]
since all prime numbers must be greater than 1. 
Step 2: Filter all numbers off of
arr
that are not prime by subjecting each element to the “trial division test” which “consists of dividing n by each integer m that is greater than 1 and less than or equal to the square root of n”. This test returnsfalse
if any number less than the element being operated on (m) produces no remainder when said element (n) is divided by it. See link below for more on this. 
Step 3: Return the sum of all remaining elements of arr using
.reduce()
.
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NOTES FOR CONTRIBUTIONS:
 DO NOT add solutions that are similar to any existing solutions. If you think it is similar but better, then try to merge (or replace) the existing similar solution.
 Add an explanation of your solution.
 Categorize the solution in one of the following categories — Basic, Intermediate and Advanced.
See
Wiki Challenge Solution Template
for reference.