# Project Euler Problem 6 Solution

First posted on 16/10/2012

### 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.

### Solution in Java

``````// Problem 6:
// Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

public class multiples {
public static void main(String[] args) {
long begin = System.currentTimeMillis();

int sum_1 = 0, sum_2 = 0, sum = 0;
int n = 100;

// Sum of the first nth integer
sum_1 = n*(n+1)/2;
sum_1 *= sum_1;

// Sum of nth squared integers
sum_2 = (n * (n+1) * (2*n+1))/6;

System.out.println(sum_1 - sum_2);

long end = System.currentTimeMillis();
System.out.println(end-begin + " ms");
}

}
``````

### Footnotes

This was my original algorithm. Brute forced.

``````for(int i = 1; i <= 100; i++){
prod = i*i;
sum_1 += prod;

before = i;
after += before;
}

sum_2 = after*after;
sum = sum_2 - sum_1;
``````

There was a quick solution for this by using Number Series we learned in elementary maths.

To find sum of the first nth integer,
$\frac{n*(n+1)}{2}$

To find sum of the squared nth integer,
$\frac{n *(n+1)*(2*n+1)}{6}$