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Peter Kim Frank
Peter Kim Frank

Posted on • Updated on

Project Euler #1 - Multiples of 3 and 5

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!

Top comments (43)

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maxart2501 profile image
Massimo Artizzu • Edited on

I'll take a mathematical approach.

The sum of the first n positive integers is n*(n + 1)/2. Multiply by k, and you get the sum of the first n multiples of k.

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:

const limit = 1000;
const m = 3, n = 5;

const mulM = Math.floor((limit - 1) / m);
const mulN = Math.floor((limit - 1) / n);

const lcm = leastCommonMultiple(m, n);
const mulLcm = Math.floor((limit - 1) / lcm);

const result = m * mulM * (mulM + 1) / 2
  + n * mulN * (mulN + 1) / 2
  - lcm * mulLcm * (mulLcm + 1) / 2;
Enter fullscreen mode Exit fullscreen mode

There, no iterations, pure math πŸ‘ And blazing fast ⚑

Using JavaScript's new support of BigInt, it's immediate even with huge numbers like 123456789012345678901234567890: it's 3556368375755728575115582031196717989244271707186887161545.

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tttaaannnggg profile image
tang

yesss the Gaussian sum is such a good tool!

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nestedsoftware profile image
Nested Software

This is very clever and very nice!

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pbouillon profile image
Pierre Bouillon

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

HAI 1.2
    CAN HAS STDIO?
    I HAS A GOAL ITZ 1000
    I HAS A TOTAL ITZ 0

    I HAS A VAR ITZ 0
    IM IN YR LOOP UPPIN YR VAR TIL BOTH SAEM VAR AN GOAL
        BOTH SAEM 0 AN MOD OF VAR AN 3
        O RLY?
            YA RLY
                TOTAL R SUM OF TOTAL AN VAR
            NO WAI
                BOTH SAEM 0 AN MOD OF VAR AN 5
                    O RLY?
                        YA RLY
                            TOTAL R SUM OF TOTAL AN VAR
                        OIC
        OIC 
    IM OUTTA YR LOOP

    VISIBLE TOTAL

KTHXBYE
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philnash profile image
Phil Nash

πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘

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pbouillon profile image
Pierre Bouillon

You can try it here

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dance2die profile image
Sung M. Kim

Answer in C# using LINQ.

Enumerable
    .Range(1, 1000)
    .Where(i => i % 3 == 0 || i % 5 == 0)
    .Sum();

Explanation

  1. Enumerable.Range generates a number between 1 & 1000
  2. Where filters records that matches a condition (It's like filter in JS)
  3. 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.

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thejoezack profile image
Joe Zack

short AND legible!

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dance2die profile image
Sung M. Kim

Thanks Joe :)

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philnash profile image
Phil Nash • Edited on

Ruby

(1...1000).select { |n| n % 3 == 0 || n % 5 == 0 }.sum
# => 233168

JavaScript

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

Array.from(Array(1000).keys())
  .filter(n => n % 3 === 0 || n % 5 === 0)
  .reduce((acc, n) => acc + n);
// => 233168
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ndc profile image
NC • Edited on

Or generate the range with the es6 spread operator:

[...Array(1000).keys()]
    .filter(n => n % 3 === 0 || n % 5 === 0)
    .reduce((acc, n) => acc + n)

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:

eval([...Array(1000).keys()].filter(n => n % 3 === 0 || n % 5 === 0).join('+'))

It's a bad practice to use eval, of course, but useful for JS code golf.

Or with lodash:

_(_.range(1, 1000)).filter(n => n % 3 === 0 || n % 5 === 0).sum()
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matheus profile image
Matheus Calegaro • Edited on

Here it is Β―\_(ツ)_/Β―
I'm looking forward for feedbacks

let sum = 0,                                        
    i = 0

  for (i = 0; i < 1000; i++) {
    if (i % 3 === 0 || i % 5 === 0) sum += i
  }

console.log('The sum is %d', sum)
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gmartigny profile image
Guillaume Martigny

Smallest nitpicking ever: declare i inside the for loop.

for (let i = 0; i < 1000; i++)

It will prevent it from leaking outside the loop.

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jacobmgevans profile image
Jacob Evans

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

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citizen428 profile image
Michael Kohl • Edited on

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

Clojure:

(defn problem1
  [n]
  (reduce + (filter #(or (= 0 (mod % 3)) (= 0 (mod % 5))) (range 1 n))))

(println (problem1 1000))

Haskell (boring):

main :: IO ()
main = print problem1

problem1 :: Integer
problem1 = sum (filter (\x -> x `mod` 3 == 0 || x `mod` 5 == 0) [1..999])

Haskell (a bit more interesting, but wasteful):

import Data.List
problem1 = sum $ nub $ [3,6..999] ++ [5,10..999]

Potion:

Number dividesBy = (x):
  self % x == 0.

sum = 0
1 to 999 (x): 
  if (x dividesBy(3) || x dividesBy(5)): sum += x
_x
(sum, "\n") join print

GNU Smalltalk:

res := ((3 to: 999) select: [:i| (i \\ 3 = 0) | (i \\ 5 = 0)]) 
  inject: 0 into: [:sum :i| sum+i].
res printNl
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rrampage profile image
Raunak Ramakrishnan

Here's one in Haskell using list comprehension:

sum [x | x <- [1..999], x `mod` 3 == 0 || x `mod` 5 == 0]
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r0f1 profile image
Florian Rohrer

Python

print(sum(i for i in range(1001) if i % 3 == 0 or i % 5 == 0))
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r0f1 profile image
Florian Rohrer

For fun: A shorter solution.

sum({*range(3, 1000, 3)} | {*range(5, 1000, 5)})
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aswathm78 profile image
Aswath KNM

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

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hanachin profile image
Seiei Miyagi

Rubyβœ¨πŸ’Žβœ¨

require "set"

multiples_of_3 = Enumerator.new {|y| x = 0; y << x while x += 3 }
multiples_of_5 = Enumerator.new {|y| x = 0; y << x while x += 5 }

puts [multiples_of_3, multiples_of_5].each_with_object(Set.new) { |e, s|
  s.merge(e.take_while(&1000.method(:>)))
}.sum
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aspittel profile image
Ali Spittel
def sum_natural_numbers_below_n(n):
    # O(N) complexity
    mult_3 = range(3, n, 3)
    mult_5 = range(5, n, 5)
    all_multiples = set(mult_3 + mult_5)
    return sum(all_multiples)

print(sum_natural_numbers_below_n(1000))
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alainvanhout profile image
Alain Van Hout

In Java

    final int sum = IntStream.range(0, 1000)
            .filter(x -> x % 3 == 0 || x % 5 == 0)
            .sum();
    System.out.println(sum);
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gabriela profile image
Gabi • Edited on

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;
}

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nektro profile image
Meghan (she/her) • Edited on
new Uint16Array(1000)
.map((v,i) => i)
.filter((v) => v % 3 == 0 || v % 5 === 0)
.reduce((ac,cv) => ac + cv)

// 233168
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ronaldoperes profile image
Ronaldo Peres

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.

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srleyva profile image
Stephen Leyva (He/Him)

Python implementation that runs in Θ(1) time

def solve(n):
    three = 3 * sum_all(math.floor(n/3))
    fives = 5 * sum_all(math.floor(n/5))
    sames = 15 * sum_all(math.floor(n/15))
    return three + fives - sames


def sum_all(n):
    return n*(n+1)/2

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_bigblind profile image
Frederik πŸ‘¨β€πŸ’»βž‘οΈπŸŒ Creemers
result = sum(x for x in range(1001) if x % 3 == 0 or x % 5 == 0)

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