If you haven’t already, go ahead and sign up at Project Euler. Once you have logged in, head on over to the Problems page. I recommend sorting the problems by ascending difficulty, as this will allow you to tackle the problems in a (more or less) gentle slope, at least at first. Let’s take a look at the first problem:

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.

So where do we get started? The key to this problem is the modulo operator. For those who are unfamiliar with the modulo operator (notated in many languages as % ), it divides two numbers and returns the remainder. So, if a number divides evenly by 3 or 5, we want to add it to a running total. So, if we were to look at this problem in language-agnostic pseudo-code we would get:

for each number between 0 and 1000 if number/3 evenly, or number/5 evenly add the number to a running total print the total to the user

With that, you should be able to complete this problem fairly quickly. If you would like to see this problem implemented in a few programming languages, hit the Read link below:

### C#

static void Main(string[] args) { int total = 0; for (int i = 0; i < 1000; i++) { if (i % 3 == 0 || i % 5 == 0) { total += i; } } Console.WriteLine(total); }

### Java

public static void main(String[] args) { int total = 0; for(int i = 0; i < 1000; i++){ if(i%3 == 0 || i%5 == 0){ total += i; } } System.out.print(total); }

### Python

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