Different solutions for Fizz Buzz in Ruby

fizzbuzz.rb

def fizz_buzz_1(max)
  arr = []
  (1..max).each do |n|
    if ((n % 3 == 0) && (n % 5 == 0))
      arr << "FizzBuzz"
    elsif (n % 3 == 0)
      arr << "Fizz"
    elsif (n % 5 == 0)
      arr << "Buzz"
    else
      arr << n
    end
  end
  return arr
end

def fizz_buzz_2(max)
  arr = []
  (1..max).each do |n|
    if (n % 3 == 0)
      if (n % 5 == 0)
        arr << "FizzBuzz"
      else
        arr << "Fizz"
      end
    elsif (n % 5 == 0)
      arr << "Buzz"
    else
      arr << n
    end
  end
  return arr
end

def fizz_buzz_3(max)
  arr = []
  (1..max).each do |n|
    text = ""
    if (n % 3 == 0)
      text << "Fizz"
    end
    if (n % 5 == 0)
      text << "Buzz"
    end
    if !((n % 3 == 0) || (n % 5 == 0))
      text = n
    end
    arr << text
  end
  return arr
end

_=$$/$$;__=_-_;@_=_+_;$_=@_+_;$__=@_+$_;$-_=$__*$_;@__=''<<$-_*($__+$_)+@_;$___=''<<$-_*$__-$__<<$-_*($__+@_)<<@__<<@__;@___=''<<$-_*$__-$_*$_<<$-_*($__+$_)-$_<<@__<<@__;(___=->{$.+=_;$><<($.%$-_==__ ?$___+@___:$.%$_==__ ?$___:$.%$__==__ ?@___:$.)<<(''<<$__*@_);$.<($__*@_)**@_?___[]:_})[]

Thanks to http://threeifbywhiskey.github.io/2014/03/05/non-alphanumeric-ruby-for-fun-and-not-much-else/

(1..100).each do |x|
  fizz = "Fizz" if (x % 3 == 0)
  buzz = "Buzz" if (x % 5 == 0)
  puts (fizz || buzz) ? fizz.to_s + buzz.to_s : x
end

class Fixnum
  def to_s
    string = ''
    string += 'fizz' if self % 3 == 0
    string += 'buzz' if self % 5 == 0
    return string unless string.empty?
    inspect
  end
end

puts (1..100).to_a

(1..100).each do |i|
if i % 5 == 0 && i % 3 == 0
puts 'FizzBuzz'
elsif i % 5 == 0
puts 'Buzz'
elsif i % 3 == 0
puts 'Fizz'
else
puts i
end

end

#FizzBuzz

(1..100).each do |number|
if (number % 3 == 0) and (number % 5 != 0)
puts "Fizz"
elsif (number % 5 == 0 ) and (number % 3 != 0)
puts "Buzz"
elsif (number % 5 == 0) and (number % 3 == 0)
puts "FizzBuzz"
else
puts number
end
end

buzzer = lambda do |i|
  map = {ff: i, tf: 'Fizz', ft: 'Buzz', tt: 'FizzBuzz'}
  map[((i%3==0).to_s[0] + (i%5==0).to_s[0]).to_sym]
end
puts (1..100).map(&buzzer)

(0..100).map{|i|i%15==0?'FizzBuzz':i%3==0?'Fizz':i%5==0?'Buzz':i}

#66 chars :)

This one requires presence from Rails (i.e., it works in rails console):

puts 1.upto(100).map { |n| "#{'Fizz' if n%3==0}#{'Buzz' if n%5==0}".presence || n }

(1..100).map{|n|({3=>['Fizz'],5=>['Buzz']}.map{|x,y|y[n%x]}.join.gsub(/^$/,n.to_s))}

@willwright82 's solution modified for less == 0:

(1..100).map do |m|
  case 0
  when m % 15 then 'FizzBuzz'
  when m % 3  then 'Fizz'
  when m % 5  then 'Buzz'
  else m
  end
end

def fizzbuzz(int)
if int % 3 == 0 && int % 5 == 0
"FizzBuzz"
elsif int % 3 == 0
"Fizz"
elsif int % 5 == 0
"Buzz"
else int % 4 == 0
nil
end
end

70 characters

(1..100).map{|x|puts x%15==0?'FizzBuzz':x%5==0?'Buzz':x%3==0?'Fizz':x}

When I thought we already had way too many solutions... !

module BeingMultiple
  refine Integer do
    def eitherFizzOrBuzz!
      return "FizzBuzz" if multiple?(3) && multiple?(5)
      return "Fizz" if multiple?(3)
      return "Buzz" if multiple?(5)
      self
    end
    
    private
    
    def multiple?(target)
      self % target == 0
    end
  end
end

using BeingMultiple

puts (1..n).map(&:eitherFizzOrBuzz!).join("\n")

require "fizzbuzz"

RSpec.describe FizzBuzz do
  describe "#compute" do
    context "when number is dvisisble by 3" do
      it "returns fizz" do
        expect(FizzBuzz.compute(6)).to eq("fizz")
      end
    end

    context "when number is dvisisble by 5" do
      it "returns buzz" do
        expect(FizzBuzz.compute(10)).to eq("buzz")
      end
    end

    context "when number is dvisisble by both 3 and 5" do
      it "returns buzz" do
        expect(FizzBuzz.compute(15)).to eq("fizzbuzz")
      end
    end
  end
end


class FizzBuzz
  def self.compute(number)
    return "fizzbuzz" if (number % 15).zero?

    if (number % 3).zero? "fizz" else "buzz" end
  end
end