Project Euler 006 – IronRuby

9 03 2011

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.

Solution

#The sum of the squares of

 #the first ten natural numbers is,

 #

 #1^2 + 2^2 + … + 10^2 = 385

 #The square of the sum of the

 #first ten natural numbers is,

 #

 #(1 + 2 + … + 10)^2 = 55^2 = 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.

 

n = 100

answer = (n*(n+1) / 2)**2 (n*(n+1)*(2*n+1)) / 6

puts answer

 

Discussion

And there’s your trick.  Incidentally, I need to create a cheat sheet for the math functions in all of these languages.  ** is not in my vocabulary.  It’s nice though.

If you have questions, leave a comment or please find me on Twitter (@azzlsoft) or email (rich@azzlsoft.com).

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