Not getting correct answers from very simple functions - c++

Question

Okay, so I'm a complete noob. I'm trying my hand at Project Euler to get better at C++. I'm doing problem #1, but I'm not getting the correct output. When I run it, I get that numTotalThree is -3, and numTotalFive is -5, and that numTotal is 0. There's something wrong with my functions, but I'm not sure what I've done wrong. How do I fix this?
#include <iostream>
using namespace std;
int main()
{
int amount = 1000;
int numOfThree = amount / 3;
int numOfFive = amount / 5;
int numTotalThree = 0;
int numTotalFive = 0;
int numTotal = numTotalThree + numTotalFive;
cout << numOfThree << endl;
cout << numOfFive << endl;
for(int i = 0; i <= numOfThree; i++)
{
numTotalThree += numTotalThree + 3;
}
cout << numTotalThree << endl;
for(int i = 0; i <= numOfFive; i++)
{
numTotalFive += numTotalFive + 5;
}
cout << numTotalFive << endl;
cout << numTotal << endl;
system("PAUSE");
return 0;
}

Answer 1

I guess you need something like this:
int sum = 0;
for (int i =0; i < 1000; ++i){
if(i % 3 == 0 || i % 5 == 0){
sum += i;
}
}
Later edit: I don't know why you want to count the numbers divisible with 3 or 5 that are less than 1000. The problem (Project Euler - Problem 1) asks for the sum of all the numbers less than 1000, divisible with 3 or 5.

Answer 2

C++ is not a functional language it's procedural - that means you have to do things in order. When you do this:
int numTotalFive = 0;
int numTotal = numTotalThree + numTotalFive;
It will be executed then and there and not again when numTotalThree and numTotalFive are updated. If you don't touch it again that's the value that will be output.

Answer 3

Here's an idea to go on:
Check how many are divisible by three by checking that the remainder %==0. Do the same for five, and then for both of them. Subtract from the total of the first two the number that is divisible by both to get an accurate answer.
int divisibleByThree=0;
int divisibleByFive=0;
int divisibleByBoth=0;
int total;
for(int i=0; i<1000; i++)
{
if (i%3==0)
divisibleByThree++;
if (i%5==0)
divisibleByFive++;
if (i%5==0) && i%5==0)
divisibleByBoth++;
}
total = divisibleByThree + divisibleByFive - divisibleByBoth;
return total;

Answer 4

Your
numTotalThree is overflowing for n in [0, 333]
3/2*(-2 + 2n)
Similarly numTotalFive for n in [0, 200]
5/2*(-2 + 2n)
So you're seeing negative values.
As other suggested you probably need to revisit your logic.
All you need is just to sum up the numbers in [0,1000] that are divisible by 3 or 5
size_t total =0;
for (size_t x =0; x < 1000; x++){
if( (x % 3 == 0) || (x % 5 == 0) ){
total += x;
}
}