Problem 6 is one of the easy ones.

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

1^(2) + 2^(2) + ... + 10^(2) = 385

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

(1 + 2 + ... + 10)^(2) = 55^(2) = 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.

So easy that the code doesn't really require a write up, there simply isn't much to this problem.

public class Euler6 { private static int sumOfTheSquares(int n) { int sum = 0; for (int i = 1; i <= n; i++) { sum += i * i; } return sum; } private static int squareOfTheSum(int n) { int sum = 0; for (int i = 1; i <= n; i++) { sum += i; } return sum * sum; } public static void main(String[] args) { int n = 100; int diff = squareOfTheSum(n) - sumOfTheSquares(n); System.out.println(diff); } }

## No comments:

Post a Comment