One Pager Cheat Sheet

• Write a method sumOfAllPrimes that takes a number n and returns the sum of all prime numbers smaller than or equal to n.
• We can use an array of length n+1 initialized with natural numbers starting from 2, and removing 0 and 1, to programmatically determine what numbers are prime.
• We can figure out the prime number candidates that are smaller than n with an O(n^2) operation by looping through every number from 2 to the current one and seeing if it is divisible.
• We can check for prime numbers efficiently by looping over the numbers below n and performing a log n time operation to check divisibility by prime numbers larger than the square root of n.
• We can use a filter that requires a number to be divisible beyond its square root in order to determine if it is a prime number, which allows us to efficiently find and sum all prime numbers.

This is our final solution.

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}

var assert = require('assert');

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function sumOfPrimes(n) {

  const numsLessThanN = Array.from({ length: n + 1 }, function (val, idx) {

    return idx;

  }).slice(2);

​

  const primes = numsLessThanN.filter((num) => {

    let primeCandidate = num - 1;

​

    while (primeCandidate > 1 && primeCandidate >= Math.sqrt(num)) {

      if (num % primeCandidate === 0) return false;

      primeCandidate--;

    }

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    return true;

  });

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  return primes.reduce((prime1, prime2) => prime1 + prime2);

}

​

try {

  assert.deepEqual(2, sumOfPrimes(2), 'sumOfPrimes(2) should equal 2');

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  console.log('PASSED: ' + '`sumOfPrimes(2)` should equal `2`');

} catch (err) {

  console.log(err);

}

​

try {

  assert.deepEqual(129, sumOfPrimes(30), 'sumOfPrimes(30) should equal 129');

​

Great job getting through this. Let's move on.

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