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

Problem 48

The series, 1¹ + 2² + 3³ + ... + 10¹⁰ = 10405071317.
Find the last ten digits of the series, 1¹ + 2² + 3³ + ... + 1000¹⁰⁰⁰.

Solution (In Ruby)

The solution to this problem becomes simple if we just directly implement with the brute force technique. It just costs two lines in Ruby and the source code is given below.

sum = (1..1000).to_a.inject(0) {|b,i| b+= i**i}.to_s
puts sum.to_s[(sum.size-10)..(sum.size)]

Hover here to see the solution.

Cheers!!
Bragaadeesh