Problem 6

Problem 6

Sixth problem from project Euler.
This one was surprisingly easy, I remember having issue with this problem back at uni while learning C#. This time around it seemed quite simple

Question:

The sum of the squares of the first ten natural numbers is,

1² + 2² + ... + 10² = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)² = 55² = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640. Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

Answer:

public class problem6 {
 public static void main(String[] args) {
  double rootSum = 0;
  double sum = 0;
  for(double i = 0; i < 101; i++){
   sum += i;
   double sqr = i * i;
   rootSum += sqr;
  }
  System.out.println((int)((sum * sum)-rootSum));
 }
}