Euler Problem 1 Python Solution

I wanted to start playing with python a bit more so I thought I’d take a look at the project Euler problems. The first problem is to find the sum of all numbers between 1 and 999 that are divisible by 3 or 5.

After solving it and looking on the forums I was kind of shocked that most solutions used loops, iterating over every number between 1 and 1000. A more efficient solution is to use the sum of an arithmetic progression. Which I would have thought is high school Maths (I still needed Wikipedia to prompt me, ho hum). Anyway Here’s my solution:

max = 999

maxthree = ((max/3)*3) + 0.0
maxfive = ((max/5)*5) + 0.0
maxfifteen = ((max/15)*15) + 0.0

multhree = ((maxthree/3)/2)*(3+maxthree)
mulfive = ((maxfive/5)/2)*(5+maxfive)
mulfifteen = ((maxfifteen/15)/2)*(15+maxfifteen)

print multhree + mulfive - mulfifteen

I’d guess about 95% of the solutions used loops.
This is my favourite solution from the forum:

Here is a solution that is reasonably efficient but it works for any list of possible factors, without being overcomplicated (in Java). It took me about 50 minutes to write it.

import java.util.*;
import java.lang.Math;

public class Euler1 {

   public static void main(String[] args) {

      // Eueler project problem 1
      // find the sum of all numbers less than N (1000) that are multiples of a given list of factors (3 and 5)
      // the program seeks a balance between speed, memory use, generality and program simplicity

      int[] factors = {3,5};
      int N = 1000;

      int sum = computeSum (factors, N);



   public static int computeSum (int[] factors, int N) {

      // efficient calculation using the fact that the pattern of multiples repeats
      // use the produce of the factors as the period length, 
      // the least common multiple would be more efficient but complicates the program

      int M = 1;
      for(int j=0; j<factors.length; j++) {
         int f = factors[j];
         M *= f;

      int k = (N-1)/M;    // number of repeated periods
      int r = (N-1)%M+1;  // remaining length;

      // use average of sum over first and last period time k, plus sum over the remaining numbers 

      int sum = (bruteSum(factors, 1, M+1)+bruteSum(factors,(k-1)*M+1,k*M+1))*k/2 + bruteSum(factors, k*M+1, N);

      return sum;

   public static int bruteSum(int[] factors, int N0, int N1) {

      // computes the answer using a straightforward but inefficient brute force method 

      int sum = 0;
      for(int i=N0; i<N1; i++) {
         boolean isAMultiple = false;
         for(int j=0; j<factors.length; j++) {
            int f = factors[j];
            if(i%f == 0) isAMultiple = true;

         if(isAMultiple) {
            sum += i;


      return sum;

Leave a Reply

Your email address will not be published.

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>