Project Euler - Problem 1

This is my first blog post, obviously.  It is a program that is designed to be a bit of a walkthrough of Project Euler's first problem.

This is the solution in c++, using a user-defined function.  Now, it's really much simpler, and you don't need the function at all. All of the stuff inside of the function can just be put into main, but I did it to show off how user-defined functions work.

// We'll be using a user defined function.
int nat_addition(int);
// This function accepts an integer as a parameter.
// The function defintion is found below main.

int main()
{
    int up_to;
   
    // Prints out a bit of a menu
    cout << "Project Euler, problem 1:";
    cout << "\nIf we list all the natural numbers below 10 that are multiples of";
    cout << "\n3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.";
    cout << "\nFind the sum of all the multiples of 3 or 5 below 1000.";
    // Asks for and accepts the upper limit of the number you want
    cout << "\n\nEnter the number up to which you would like to find";
    cout << "\nthe sum of the multiples of 3 or 5: ";
    cin >> up_to;
    cout << "\n" << nat_addition(up_to);
   
    system("PAUSE");
    return 0;
}


// nat_addition finds the sum of all of the multiples of 3 or 5
// below a specified number.
int nat_addition(int Below)
{
    int sum = 0; // Note that sum is initalized to 0
                 // This is to prevent sum from having a value
                 // already stored in it that we don't want.
   
    for (int i = 1; i <= Below; i++) { // This for loop
        if (i%3 == 0 || i%5 == 0) {    // is designed to loop
            sum += i;                // through every number between
        }                        // one and the integer specified. If the
    }                        // number is divisible by 3 or 5, it is added
                // to sum, which is then returned.
    return sum;
};