computeSum(integers)

JavaScript
#1

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

#2

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.

#3

Thanks for your prompt response, I tried as directed by you, see code below, but it is still failing. Regards. please help:
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]);

#4

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

#5

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.

#6

How is he starting at 6?

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

#7

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

#8

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

#9

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

1 Like
#10

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

#11

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

#12

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

#13

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

#14

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).

#15

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.

#16

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.

#17

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;
}

#18

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);
#19

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);

#20

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