# Project Euler, problem 1, multiple factors

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.

To solve this in a brute force manner, we can count up to 1000, for each number we check to see if it divides into 3 and 5 leaving no remainder. If it does, they we add that number into our running total.

Heres my solution using Java.

```
public class Problem1 {
public static void main(String[] args) {
System.out.println(solve(1000));
}
public static int solve(final int subject) {
int sum = 0;
for (int i = 0; i < subject; i++) {
if ((i % 3 == 0) || (i % 5 == 0)) {
sum += i;
}
}
return sum;
}
}
```

The answer is 233168