SumPrimes Nested Loops Solution - Is my logic flawed?

JavaScript
1

I’m not sure what is wrong with my code. Am I on the right track?
Sorry I’m still a novice with JS.


function sumPrimes(num) {
var sum = 0;
var currentVal = 2;
while (currentVal <= num) {
  for (var i = 2; i < currentVal; i++) {
    if (currentVal % i == 0) {
      continue;
  } else {
    sum += currentVal;
    }
  }
  currentVal++;
}
return sum;
}

console.log(sumPrimes(10));

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Challenge: Sum All Primes

Link to the challenge:

2

Can you describe, in words, what this if-else is supposed to do?

3

Its supposed to mean:
“If the current value divided by the value of i has no remainder then it is not a prime number, so then move onto the next value (currentVal += 1).
Otherwise, if it has a remainder, then add to the sum total.”

Is the problem that I am adding to the sum values that aren’t actually prime numbers?

4

Two problems with that description:

  1. having no remainder means that i divides evenly into currentVal

  2. a number greater than 2 is prime if no values other than 1 and itself divide evenly into it

5
  1. Right, so what my code currently says is this:
    “If i divides evenly into currentVal then skip to i += 1.
    Otherwise, if i does NOT divide evenly into currentVal, add currentVal to sum.”

But what I realize now is I’m not checking all possible values of i < currentVal before adding currentVal to sum. I’m adding currentVal as soon as one particular value of i does not divide evenly without checking the rest of possible values.
Is this my error?

6

Exactly. You are currently adding currentVal to the sum for each number less than currentVal that does not divide into currentVal. So you have the right piece being combined in a way that isn’t quite what you want.

1 Like
7

Hey Jeremy. So I stuck with it, tinkered with my code and I got it to work!
I incorporated a boolean flag.

function sumPrimes(num) {
  var sum = 0;
  var currentVal = 2;
  var flag = true;
  while (currentVal <= num) {
    for (var i = 2; i < currentVal; i++) {
      if (currentVal % i == 0) {
        flag = false;
        break;
    } else {
        flag = true;
      }
    }
  if (flag) {
    sum += currentVal;
  }
  currentVal++;
}
  return sum;
}
console.log(sumPrimes(10));
1 Like
8

Nice work figuring it out. As a bonus question, how can you change this one line to make it faster?

1 Like
9

I only know this from peeking at the provided solutions:

for (var i = 2; i <= Math.sqrt(currentVal); i++) {

The logic is that the square root of a number is the largest unique divisor possible.

1 Like
10

I like those solutions - I think they present three interesting ways to look at the problem.

1 Like
11

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