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).
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") } }
