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Project Euler #1 - Multiples of 3 and 5

Peter Kim Frank on June 11, 2018

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 th...

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Massimo Artizzu • Jun 11 '18 • Edited

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;

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.

Nested Software • Jun 12 '18

This is very clever and very nice!

mari tang • May 29 '19

yesss the Gaussian sum is such a good tool!

Pierre Bouillon • Jun 12 '18

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

Phil Nash • Jun 12 '18

👏👏👏👏👏

Pierre Bouillon • Jun 12 '18

You can try it here

Sung M. Kim • Jun 12 '18

Answer in C# using LINQ.

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

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.

Joe Zack • Jun 12 '18

short AND legible!

Sung M. Kim • Jun 12 '18

Thanks Joe :)

Phil Nash • Jun 12 '18 • Edited

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

NC • Jun 13 '18 • Edited

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()

Matheus Calegaro • Jun 12 '18 • Edited

Here it is ¯\_(ツ)_/¯
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)

Guillaume Martigny • Jun 13 '18

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.

Jacob Evans • Sep 7 '19

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

Florian Rohrer • Jun 11 '18

Python

print(sum(i for i in range(1001) if i % 3 == 0 or i % 5 == 0))

Florian Rohrer • Jun 11 '18

For fun: A shorter solution.

sum({*range(3, 1000, 3)} | {*range(5, 1000, 5)})

Aswath KNM • Jun 12 '18

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

Ali Spittel • May 28 '19

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))

Seiei Miyagi • Jun 21 '18

Ruby✨💎✨

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

Alain Van Hout • Jun 12 '18

In Java

    final int sum = IntStream.range(0, 1000)
            .filter(x -> x % 3 == 0 || x % 5 == 0)
            .sum();
    System.out.println(sum);

Omkar Ajnadkar • Jun 13 '18 • Edited

Python

sum = 0
for i in range(0,1000):
    if i % 3 == 0 or i % 5 == 0:
        sum = sum + i
print(sum)

or Python One-Liner

print(sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0))

Gabi • Jun 3 '19 • Edited

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;
}

Meghan (she/her) • Jun 11 '18 • Edited

new Uint16Array(1000)
.map((v,i) => i)
.filter((v) => v % 3 == 0 || v % 5 === 0)
.reduce((ac,cv) => ac + cv)

// 233168

Ronaldo Peres • Oct 30 '18

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.

Stephen Leyva (He/Him) • Jun 13 '18

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

Frederik 👨‍💻➡️🌐 Creemers • Jun 11 '18

result = sum(x for x in range(1001) if x % 3 == 0 or x % 5 == 0)

Daniel Bruynooghe • May 22 '20 • Edited

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)

rahy-hub • Apr 7 '20 • Edited

/*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

LikeLocusts • Jun 12 '18

Scala:

def sumOfMultiples(max: Int): Int = {
    if (max <= 0) 0
    else if (max % 3 == 0 || max % 5 == 0) max + sumOfMultiples(max - 1)
    else sumOfMultiples(max - 1)
  }

Big-Zude • Jun 4 '19

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
var multi5=((5*b)(b+1))/2
var multi15=((15*c)*(c+1))/2

var sum=(multi3+multi5)-multi15
return sum
}
console.log(multiple())

Muhammad • Jun 12 '18

I did that a while ago in C++

Jake Varness • Jun 12 '18

Here it is in Go:

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

Raunak Ramakrishnan • Jun 12 '18

Here's one in Haskell using list comprehension:

sum [x | x <- [1..999], x `mod` 3 == 0 || x `mod` 5 == 0]

sawankrr • Jan 22 '20

Java

 Long sum = LongStream.range(1, 100)
                    .filter(i -> i % 3 == 0 || i % 5 == 0)
                    .sum();
    System.out.println(sum);

Erhan Kılıç • Jun 13 '18

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...

Sai gowtham • Jun 13 '18 • Edited

let sum=0;
let start=999;

 while(start){
    if(start % 3 === 0 || start % 5===0){
       sum +=start;
    }  
   start--;  
 }

console.log(sum);

Daniel Bruynooghe • May 22 '20 • Edited

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

Daniel Bruynooghe • May 22 '20 • Edited

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.

JuriTuuti • Jun 20 '21

Julia:
sum([n for n=1:999 if n%5==0 || n%3==0])