Exercise 8 Sum of multiples of 3 and 5

Hey everyone,
When I was following the JavaScript exercise 8 about sum of multiples of 3 and 5 I think I found a bud in the code Mosh showed in the video. I don’t know if this is not checked by him by mistake or if it was intentionally.

The exercise is to create a function that adds the multiples of 3 and 5 under a limit number:
For example
If the limit number is 10, the multiples of 3 and 5 are: 3, 6, 9 and 5, 10.
Then he would add those and the result is 33.

But if the limit number is 15, the multiples are: 3, 6, 9, 12, 15 and 5, 10, 15.
In this case 15 is double in the sequence and I realised in this code the result was 60 instead of 75 because he didn’t add the doubled 15.

The code he used:


function sum(limit) {
let sum = 0;
for (let i = 0; i <= limit; i++) if (i % 3 === 0 || i % 5 === 0) sum += i;

return sum;

I did try to code a function that would add numbers if they occur in both the multiples of 3 and 5 but its very long and since I’m really new to coding its probably not efficient, but it does work :wink:

I was hoping if someone that knows more could also try to code this problem so I could see how you would solve this problem on a higher level than me XD

btw this is what I came up with:


function sum(limit) {
i = 0;
let sumMultipliedByThree = 0;
let sumOfThree = 0;

while (sumMultipliedByThree <= limit) {
sumOfThree += sumMultipliedByThree;
sumMultipliedByThree = 3 * i;

i = 0;
let sumMultipliedByFive = 0;
let sumOfFive = 0;

while (sumMultipliedByFive <= limit) {
sumOfFive += sumMultipliedByFive;
sumMultipliedByFive = 5 * i;
console.log(sumOfThree + sumOfFive);

Thanks for the help in advance, this is really just a curiosity question for people who like to answer

This must be on purpose because the way the for loop works i will have an eligible value once. So doubles do not apply


function sum(limit) {
	let sum = 0;
	for (let i = 0; i <= limit; i++) 
		if (i % 3 === 0 || i % 5 === 0)
			sum += i;

	return sum;

That said the exercice seems inspired by the FizzBuzz problem which is a bit different.

Your code is complex, verbose and uses multiple loops.
I’d refactor like this


function sum(limit) {
	let sumBy3 = 0;
	let sumBy5 = 0;
	for (let i = 0;  i <= limit; i++) {
		if (i % 3 === 0) sumBy3 += i;
		if (i % 5 === 0) sumBy5 += i;

  return sumBy3 + sumBy5;


1 Like

thanks for the response, love to see how people write code that that seem so easy when its done and seem so hard when you have to write it XD

Hi - I was reading your answer to the op, and was wondering if you could clarify something for me.
The variable statement let sum; does not work, and the return is NaN. Changing to let sum = 0; and the loop works fine. I thought - since javascript is dynamically typed, that the sum += i; would result in sum being a number by virtue of the assignment. I was wrong, but don’t understand why. Thanks.

let sum; will initialize sum as undefined. If sum is undefined, sum += 1 does not initialize sum to 0, then add 1 to it. It just adds 1 to sum, and undefined + 1 = NaN. So I think the main point is that the += operator does not do any initialized of variables.

for example:

sum += 1;

The above returns an error: “sum is not defined”. So you must declare sum, and since you intend to add numbers to it, you had best initialize it to 0 before you use it this way.

Thank you so much. I thought as much based on the result, but could not find a specific reference. New note to self - initialization never hurt. :slight_smile: