Apr 9 2009

Project Euler Problem 48 Solution

by Ash • Project Euler • Tags: Project Euler, Ruby 2

The series, 1^(1) + 2^(2) + 3^(3) + ... + 10^(10) = 10405071317.

Find the last ten digits of the series, 1^(1) + 2^(2) + 3^(3) + ... + 1000^(1000).

My solution in Ruby:

sum = 0
for i in 1..1000 do
  sum += i**i
end
str = sum.to_s
 
puts str[str.length - 10,str.length]

s = 0
(1..1000).inject { |s, x| s + x ** x } % (10 ** 10)
str = s.to_s
puts str[str.length - 10,str.length]