Counting every 5th for loop

Question

I'm having a hard time with this problem here:

Write a function named sumEvery5th that accepts a non-negative integer n and returns the sum of the integers divisible by 5 from 1 to n , including n itself. Use a for loop.

This is what I have so far:

var sumEvery5th = function(n){

    let sum = 0;
    for(let i = 1; n % 5 == 0; i++){
        sum+ i
    };
    return sum;
}

I feel like I'm really close. Thanks for the help in advance.

Answer 1

You can start your loop at 5 , since 1, 2... are not divisible by 5. And instead of i++ , you can directly go 5 by 5, until i is greater than n :

 var sumEvery5th = function(n) { let sum = 0; for (let i = 5; i <= n; i += 5) { sum += i; }; return sum; } console.log(sumEvery5th(10)); // 5 + 10 = 15 console.log(sumEvery5th(18)); // 5 + 10 + 15 = 30

Answer 2

Pure maths, no loop solution:

 var sumEvery5th = function(n) { n = Math.floor(n / 5); return 5 * n * (n + 1) / 2; } console.log(sumEvery5th(10)); // 5 + 10 = 15 console.log(sumEvery5th(18)); console.log(sumEvery5th(20));

Answer 3

First I think you should understand how a for loop works.

var sumEvery5th = function(n){
    let sum = 0;
    for(let i = 1; n % 5 == 0; i++){
        sum+ i
    };
    return sum;
}

What you are doing, step by step, is:

1. Declaring a variable i with value 1.
2. Dividing n by 5 and taking the remainder value and comparing it with 0. In case it's true, you are skipping the code block inside the for and moving towards the return sum; line.
3. (In case you haven't skipped the code block in step 2) Run the code block with the new i value.
4. (In case you haven't skipped the code block in step 2) Incrementing the i value.
5. Go back to step 2.

Usually your for condition will depend in the variable declared in step 1. What you want to do is run the for code block n times. For that, you need to change your condition from n % 5 == 0 to i <= n . This will make sure to run the code block while your i is less or equal than n , starting with a value of 1.

Now, inside your code block you add your divisible by 5 logic, checking against i value.

for(let i = 1; i <= n; i++){
    if (i%5 == 0) sum += i;
};

Now let's say I called sumEvery5th(5).

1. Declare a variable i with value 1.
2. Check if i (1) is less than or equal n (5).
3. Go inside the code block.
4. Check if i%5 is 0.
5. It's not.
6. Increment i, now i = 2.
7. Check if i (2) is less than or equal n (5). ...And so on, until i = 6, and in that case the code block is skipped and the program will continue its course.

Ps.: There are ways to improve the performance of this algorithm, but now that you understand your for loop a bit better I'll leave it to you:)

Answer 4

var sumEvery5th = function(n){

    let sum = 0;
    for(let i = 1; i <= n; i++){
        if (i % 5 === 0) {
            sum += i;
        }
    }
    return sum;
}