I thought it would be fun to create a thread where the community could solve a problem from Project Euler. This is **Problem #1**:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

Look forward to seeing solutions in your language of choice!

## Discussion (42)

I'll take a mathematical approach.

The sum of the first

npositive integers isn*(n+ 1)/2. Multiply byk, and you get the sum of the firstnmultiples ofk.There are 333 multiples of 3 below 1000, and 199 multiples of 5, so we get 166833 and 99500, respectively. But we can't just sum them, as we'd count the multiples of 15 twice (15 is their least common multiple), so we have to subtract those once (total: 33165).

Result: 233168

In general, let's use JavaScript for example:

There, no iterations, pure math 👍 And

blazing fast⚡Using JavaScript's new support of

`BigInt`

, it's immediate even with huge numbers like 123456789012345678901234567890: it's 3556368375755728575115582031196717989244271707186887161545.This is very clever and very nice!

yesss the Gaussian sum is such a good tool!

I did this in LOLCODE (first program that I'm writing with it, probably not the best implementation)

👏👏👏👏👏

You can try it here

## Ruby

## JavaScript

It's a shame getting a range and a sum isn't quite as easy in JS.

Or generate the range with the es6 spread operator:

There's also a nasty (but shorter)

`eval`

hack to use in place of reduce. You can use`.join(+)`

to convert the array into a string containing each number joined by the '+' sign, then evaluate that string as if it's a JavaScript expression to get the sum:It's a bad practice to use eval, of course, but useful for JS code golf.

Or with lodash:

Answer in C# using LINQ.

Explanation

`Enumerable.Range`

generates a number between 1 & 1000`Where`

filters records that matches a condition (It's like`filter`

in JS)`Sum`

is a convenience method for summing up a sequence (instead of doing`Aggregate((acc, n) => acc + n))`

, which is equivalent to`reduce`

in JS)Source & Tests on Github.

short AND legible!

Thanks Joe :)

At last count I did that exercise in 11 different languages, so maybe I'll post some of the less common ones.

Clojure:

Haskell (boring):

Haskell (a bit more interesting, but wasteful):

Potion:

GNU Smalltalk:

Here's one in Haskell using list comprehension:

Here it is ¯\_(ツ)_/¯

I'm looking forward for feedbacks

Smallest nitpicking ever: declare

`i`

inside the`for`

loop.It will prevent it from leaking outside the loop.

I wouldn't even call that nitpicking, that is solid best practice advice to avoid memory leaks.

Python

For fun: A shorter solution.

Great solution. Since I lost touch of python it was a little difficult. Now I understand.

Ruby✨💎✨

Hi, I am adding my first thought on this in Java.

private int sum(int inputValue) {

int sum = 0;

for (int i = 1; i < inputValue; i++) {

if (i % 5 == 0 || i % 3 == 0) {

sum += i;

}

}

return sum;

}

Python implementation that runs in Θ(1) time

In Java

Hi,

I am trying to solve this one, and already solved.

I am going to ask, what is the sum of all the multiples of 3 or 5 below 100000??

I am trying to do this but is giving me a timeout, when using loops.

## Python

or Python One-Liner

/*Multiples of 3 and 5

Problem 1

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000. */

## include

using namespace std;

int main()

{

int sum=0;

for(int i=1;i<1000;i++)

{

if(i%3==0 ||i%5==0)

{cout<<i<<" ";

sum+=i;}

}

cout<<"\n\n"<<"the sum of all the multiples of 3 or 5 below 1000 = "<<sum<<"\n";

return 0;

}

C++ >> Output >> 233168

Scala:

function multiple(){

var a=Math.trunc(999/3)

var b=Math.trunc(999/5)

var c=Math.trunc(999/15)

var multi3=((3*a)

(a+1))/2(b+1))/2var multi5=((5*b)

var multi15=((15*c)*(c+1))/2

var sum=(multi3+multi5)-multi15

return sum

}

console.log(multiple())

q){sum x where 0=min x mod/:(3,5)}til 1000

k) {+/ &: 0=min x {x-y*x div y}/:(3,5)}(!)1000

or mix and match without lambdas is even shorter and more obfuscated

q)sum (&:) 0=min ((!)1000) mod/: (3,5)

I did that a while ago in C++

Here it is in Go:

github.com/jvarness/euler/blob/mas...

JavaI started to solve problems but, well, you know little time so I couldn't continue :) I created a repository at github.

github.com/erhankilic/project-eule...

q)sum (&) 0=min mod/:[; 3 5](!)1000

k) +/&0=min{x-y*x div y}/:[;3 5](!)1000

For a change my k solution is more wordy because the mod function is rather long. Said that, the q above has k in it as well. In pure q it would read

q)sum where 0=min mod/:[;3 5]til 1000

terse and somewhat obfuscated, though the main elements are recognizable, i.e. create the list up to 1000, compute the remainders when dividing by 3 and by 5 and sum across all where either remainder is 0.

Hey, Why this doesn't work for 1000?

function multiplesOf3and5(number) {

var sum=0;

while(number>0){

if(number%3===0 || number%5===0){

sum+=number

}

number--

}

return sum;

}