From: "Eric I." <rubytraining@...>
Date: 2007-06-04T05:40:32+09:00
Subject: Re: FizzBuzz (#126) [SOLUTION]

Here's my solution.  It makes heavy use of the Ruby idiom of using the
&& and || operators to perform conditionals because a) they are short-
circuited and because b) && returns the right-hand-side when the left-
hand-side is a true value.

====
class Integer; def factor? n; self % n == 0; end; end

puts (1..100).map { |i| i.factor?(15)&&"FizzBuzz" || i.factor?
(3)&&"Fizz" || i.factor?(5)&&"Buzz" || i }
====

Eric

Are you interested in on-site Ruby training that uses well-designed,
real-world, hands-on exercises? http://LearnRuby.com