Peter Kim Frank

Posted on Jun 11, 2018 • Edited on May 28, 2019

Project Euler #1 - Multiples of 3 and 5

#projecteuler #challenge

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!

Latest comments (40)

JuriTuuti • Jun 20 '21

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

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.

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

sawankrr • Jan 22 '20

Java

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

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())

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;
}

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))