# computeSum(integers)

Compute the sum of all integers that are multiples of 5, from 1 to 100

Have you tried writing code? If you can’t think of anything, here’s what you have to do:
Initalize a variable outside a `for` loop.
In the loop, create a variable that starts at `5` and ends at `100`, stepping by `5` as you go.
Inside the loop, add the number to the variable that you have initialized.

function computeSum(integers) {
var sum = 0;

``````for (var x = 5; x < integers.length; x += 5) {
if (x % 5 === 0 && x > 0) {
sum += x;
//console.log(x);
//console.log(sum);
}
}

return sum;
``````

}
computeSum([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100]);

Arrays are 0 indexed, you’re starting at 6, and you don’t seem to be getting the actual values from the array: either you take a number, in which case you don’t pass in an array, or you pass in an array, in which case you need to get the array values

You do not need a large array like you have made. All you need is the maximum number (cap) and it will work on its own.
Also, you do not need to check `x % 5 === 0 && x > 0` because you’re starting at `x = 5` (so it’s always greater than 0) already, and you’re incrementing by 5 (so you don’t have to check if `x` is divisible by 5).

Try removing the `if` and replace `[1, 2, 3, 4, ..., 100]` with `100`. Also, make a change to your `for` loop, replacing `integers.length` with `integers`.

How is he starting at 6?

`for (var x = 5; x < integers.length; x += 5)`

He’s passing an array in, and I got very confused by that. Ugh

I know. There’s no need for him to do that. Looks like a newbie mistake. (I almost cringe.)

He does doesn’t need a loop either, this is doable as just a basic calculation

1 Like

yes I am a beginner, what do I do. regards.

From context, I can assume that he’s not that experienced in computers, so I’d leave it at that and let him learn more before I give him a formula to do it lol

Oh I get that, it’s maths though, not to do with computers

n-Karta, pls help with the formula as I am still stuck, even after carrying out the correction please.

Okay. I’ll post the better solution then.

When you’re adding up multiples, it looks like this:

`5 + 10 + 15 + 20`

If you notice, the first and last numbers add up to `25`. So do the 2nd numbers from the left and right, `10` and `15`. This is true for any set of multiples arranged in a correct order (although the sum may vary). Try it.
For instance, here are the first 6 multiples of `4`.
`4, 8, 12, 16, 20, 24`

There’s a sort of symmetrical pattern here. The first and last numbers add up to a `28`. The second and second last numbers add up to `28`. So do the third and third last numbers.

This is where a pattern emerges. All you have to do is find `first number + last number` and multiply it by `number of numbers / 2`.

In your case, `(5 + max) * (max / 5) / 2`. (`max / 5`, since `max` is the largest multiple you want, in your case, `100`).

This is actually a play on arithmetic means, but I’ll let you figure that out on your own. Yes, this also works for odd numbers of multiples (if you divide an odd number by 2 you get a `.5` which evens stuff out).

Thanks for your valuable time, I am a newbie as discovered but wanting to learn. I read and comprehended your response but can not turn it into code that works, having tried over and over again. Please help.

If `computeSum` takes a number, then if you remove `.length` from your loop, and remove the `if` block, and just do `sum += x`, that will add every fifth number up to but not including the max number.

Alternatively, just do a calculation rather than looping over anything, as @n-kartha described.

If `computeSum` takes an array of numbers then you need to access the array, so leave `.length` there, start `x` at 4, not 5, and `sum += integers[x]`. Again, you don’t need the `if`.

I tried it like this(below) and it still return error of Must be a number greater than 0
function computeSum(integers){
var sum = 0;
for(var x = 4; x < integers.length; x+=5){
sum += integers[x] ;
//console.log(x);
//console.log(sum);

}

return sum;
}

Ugh. I’ll just give you the code for the loop.

``````function computeSum(n) { // renamed integers to n for brevity
let sum = 0;

for (let i = 5; i <= n; i++) { // start at 5, and loop until i <= n, instead of n.length
sum += i; // this is what I've been trying to tell you. Just add i.
}

return sum; // return the sum of numbers
}
``````

You would use it like this:

``````computeSum(100);
``````

I tried it like you said(below) but it still complains of Number must be greater than 0.:

function computeSum(n) { // renamed integers to n for brevity
let sum = 0;
for (let i = 5; i <= n; i++) { // start at 5, and loop until i <= n, instead of n.length
sum += i; // this is what I’ve been trying to tell you. Just add i.
console.log(i);
console.log(sum);
}
return n
}
computeSum(100);

Where is this challenge you’re trying to do, because the error you’re getting makes no sense without some context?