Project Euler | Problem 6 | Sum square difference

Problem Description :

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?

Project Euler Problem 6

Concept :

This is a very simple problem. We have to apply following two properties to solve this problem mathematically.

Sum of first N natural numbers is N(N+1)/2.

Sum of square of first N natural numbers is N(N+1)(2N+1)/6

Java Program :

package com.javamultiplex.projecteuler; /** * * * @author Rohit Agarwal * @category Project Euler Problems * @problem Sum square difference * */ public class Problem6 { public static void main(String[] args) { long limit = 100; // 1^2+2^2+3^2+.....+n^2=(n*(n+1)*(2n+1))/6 long sumOfSquaresOfNaturalNumbres = (limit * (limit + 1) * ((2 * limit) + 1)) / 6; // 1+2+3+4+...+n=(n*(n+1))/2 long sumOfNaturalNumbers = (limit * (limit + 1)) / 2; long squareOfSumOfNaturalNumbers = (long) Math.pow(sumOfNaturalNumbers, 2); long difference = Math.abs(squareOfSumOfNaturalNumbers - sumOfSquaresOfNaturalNumbres); System.out.println("The sum square difference is : " + difference); } }

Output :

Output - Project Euler Problem 6

References :

https://projecteuler.net/problem=6

http://www.9math.com/book/sum-first-n-natural-numbers

http://www.9math.com/book/sum-squares-first-n-natural-numbers

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Tags : Project Euler Problem 6 solution in Java, Sum square difference, Mathematics problems, Sum of first n natural number, sum of square of first n natural numbers.

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