Project Euler #6 - Sum Square Difference

Continuing the wonderful community solutions to Project Euler.

This is Problem 6, sum square difference.

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.

Comments

[Sagnik Ray]

public class ProjectEuler{
static int n = 100;
public static void main(String []args){
System.out.println(differencehelper());

}

static int sumsquare(int n){

    int sum = 0 ;

    sum = (n*(n+1)(2*n+1))/6;

    return sum;

}

static int squaresum(int n){

   int sum  = 0;

   sum = (n(n+1))/2;

   return (sum *sum);

}

static int differencehelper(){

    return squaresum(n) - sumsquare(n);

}

}

Easy Java Solution using Gauss' Formula Hope you liked it ;)

[Ademola John]

Here is my Javascript Solution

function sumSquareDifference(n) {
  // Good luck!
  let a = 0,b=0,c=0;
  for(let i = 1;i<=n;i++){
    a+= (Math.pow(i,2));
    b+=i
  }
c=Math.pow(b,2)-a

console.log(a);
console.log(b);
  return c;
}

[rahy-hub]

my code with C++

/*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.*/

include

using namespace std;
int square_sum(int s);

int main()
{
int limit=100 ,sum=0 ,sumQ=0;
for(int i=1;i<=limit;i++)
{
sumQ=sumQ+(i*i);
sum =sum+i;
}

cout<<"the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum is \n ";
cout<< square_sum(sum)-sumQ ;  // without function : cout<<sum*sum-sumQ
return 0;

}

int square_sum(int s)
{
return s*s;
}
Output >> 25164150

[nurchikgit]

not bad)))

[Mahmoud Bayazid]

new_list = []

for i in range (1,101):
new_list.append(i*2)
a = sum(new_list)
b = sum(range(1,101))*2
print(b-a)

[Mahmoud Bayazid]

new_list = []

for i in range (1,101):
new_list.append(i^2)
a = sum(new_list)
b = sum(range(1,101))^2
print(b-a)

[hishamkhr]

hi
i try this:
sum of numbers= n(n+1)/2
then i get the square.
after that i found an equation about sum of squares:
(2*n^3 +3*n^2+n )/6
but i didn't get right result !!!!
so where i lost the control ?
thanks

[Olivier Picault]

From ProjectEuler:

Please do not deprive others of going through the same process by publishing your solution outside of Project Euler; for example, on other websites, forums, blogs, or any public repositories (e.g. GitHub), et cetera. Members found to be spoiling problems will have their accounts locked.

:)

[Oleksii Filonenko]

Elixir:

(1..100 |> Enum.map(&:math.pow(&1, 2)) |> Enum.sum()) -
  (1..100 |> Enum.sum() |> :math.pow(2))

[David Skinner]

Go

goplay.space/#rQ0XekiCknZ

// func SumSquareDifference receives the max range
// and returns the difference of the sumSquare and the squareSum.
func SumSquareDifference(x int) int {
    sum := 0
    sumSquare := 0

    for i := 1; i <= x; i++ {
        sum += i
        sumSquare += i * i
    }

    squareSum := sum * sum

    return squareSum - sumSquare
}

You can see the complete program on the playground link.

• I have included a block that confirms that my function provides the expected response.
• I then start a timer so I can print the elapsed time to run the function.
• After printing the results, I include a comment defining the Output, If the output changes it will generate an error on my computer.
• The language is Go but it is more a C++ style, I am not the best Go programmer.
• Code needs to be readable, compilers can do the optimization.

[Carlos E. Barboza]

Swift:

func sumSquareDifference(range: Int) -> Int {
    var sumOfSquares = 0
    var squareOfSum = 0
    for number in 1..<range+1 { //Is there a better way? Other than range+1?
        sumOfSquares += (number*number)
        squareOfSum += number
    }
    squareOfSum = squareOfSum*squareOfSum
    return (squareOfSum - sumOfSquares)
}

print(sumSquareDifference(range:100))