Exercise 8 Sum of multiples of 3 and 5

JavaScript
1

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:

console.log(sum(15));

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:

sum(15);

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

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

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

while (sumMultipliedByFive <= limit) {
i++;
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

2

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

console.log(sum(15));

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

console.log(sum(15));

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;
}

Regards

1 Like
3

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

4

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.

5

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;
console.log(sum);

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.

6

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: