Project Euler Answer Codes 6-10

# Problem 6

Sum square difference

I luckily came across the formula for sum of the squares of consecutive numbers which goes like this 1^2 + 2^2 + 3^2 + ... + (2n)^2 = (n(2n+1) (4n+1))/3 Then using the equation for fo sum of a series, you can get the equation for square of the sum 1 + 2 + .... + m = (m/2)(1+m) (1 + 2 + .. + m)^2 = ((m/2)(1=m))^2 Since m=2n, we can substitute and write down the difference of the 2 equations as difference = (12(n^4) + 4(n^3) - 3(n^2) - n)/3 and here n = 50

# Problem 7

10001st prime

In matlab just run factor(n) on a loop of odd numbers and increment a counter everytime you find a number with only one factor(i.e. it is a prime number). Stop when 10001st prime number is found(counter reaches 10001).

# Problem 8

Largest product in a series

Simplest and laziest way

y = ['73167176531330624919225119674426574742355349194934' ...

'96983520312774506326239578318016984801869478851843' ...

'85861560789112949495459501737958331952853208805511' ...

'12540698747158523863050715693290963295227443043557' ...

'66896648950445244523161731856403098711121722383113' ...

'62229893423380308135336276614282806444486645238749' ...

'30358907296290491560440772390713810515859307960866' ...

'70172427121883998797908792274921901699720888093776' ...

'65727333001053367881220235421809751254540594752243' ...

'52584907711670556013604839586446706324415722155397' ...

'53697817977846174064955149290862569321978468622482' ...

'83972241375657056057490261407972968652414535100474' ...

'82166370484403199890008895243450658541227588666881' ...

'16427171479924442928230863465674813919123162824586' ...

'17866458359124566529476545682848912883142607690042' ...

'24219022671055626321111109370544217506941658960408' ...

'07198403850962455444362981230987879927244284909188' ...

'84580156166097919133875499200524063689912560717606' ...

'05886116467109405077541002256983155200055935729725' ...

'71636269561882670428252483600823257530420752963450' ];

n = 1;

length(y)

max = 1;

while n < 997

     prod = str2num(y(n)) * str2num(y(n+1)) * str2num(y(n+2)) *  str2num(y(n+3)) * str2num(y(n+4)); 

     if max < prod

         max = prod;

     end

     n = n +1;

end

max

# Problem 9
No idea when this went



# Problem 10

Sum of primes below 2e6

Scala code: 

/**  * Created by Administrator on 6/6/2017.  */object Problem10 {

  //find the sum of primes below 2 million
  def main(args: Array[String]): Unit = {

    var total:Long = 0;
    var n = 0;
    var break: Boolean = false;

    for (x <- (2 to 2000000)) {

      n = (math.sqrt(x.toDouble)).toInt;
      
      break = false;

      while (n > 1 && !break) {
        if (x % (n) == 0) {
          break = true;
        }
        n -= 1;
      }

      if (!break) {
        total += x;
      }


    }
    println(s"total : $total")
  }
}