Problem 006 : Sum Square Difference

Sum square difference

Problem 6

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.

                                                                                                           https://projecteuler.net/problem=6

Problem 6 is a really simple problem if you know how to use for loop.

However, just like problem 1, I used formula to solve this problem, instead of for loop.

The formula for the sum of first n integers is:

n * (n + 1) / 2

The formula for the sum of first n squares is:

n * (n + 1) * (2n + 1) / 6

Therefore, the sum of the squares is:

int sumsquare = (NUM*(NUM+1)*(2*NUM+1))/6; //sum of the squares

and the square of the sum is:

int squaresum = (int)Math.pow((NUM)*(NUM+1)/2, 2);  //square of the sum

Answer is 25164150

Execution time is 0.8 ms, or 0.0008 seconds

Source code: https://github.com/Ainodyne/Project-Euler/blob/master/Problem006.java