# Project Euler #1 - Multiples of 3 and 5

### Peter Kim Frank Jun 11 '18 γ»1 min read

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!

dev.to is where software developers read, write, and level up.
Sign Up Now

(open source and free forever β€οΈ)

(open source and free forever β€οΈ)

Classic DEV Post from Aug 14 '18

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 π And

blazing fastβ‘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!

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

πππππ

You can try it here

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 :)

## 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:

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:

Python

For fun: A shorter solution.

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

Here it is Β―\_(γ)_/Β―

I'm looking forward for feedbacks

Smallest nitpicking ever: declare

`i`

inside the`for`

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

Rubyβ¨πβ¨

Python implementation that runs in Ξ(1) time

In Java

## Python

or Python One-Liner

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.

Scala:

I did that a while ago in C++

Here it is in Go:

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

I 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...