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.
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]);
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
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?