PROJECT EULER SOLUTION # 6

Solution to problem number 6 of Project Euler.

Question # 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.

Solution # 6

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#include<stdio.h>

#include<conio.h>

#include<time.h>

int main()

{

                long n=100,a=0,b=0;

                int i;

               

                for(i=1;i<=n;i++)

                                a=a+i;

                a=a*a;

                for(i=1;i<=n;i++)

                                b=b+(i*i);

                printf("Answer = %ld\n",(a-b));

                printf("EXECUTION TIME = %f",clock()/(float)CLK_TCK);

                system("pause");

}

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