solution for words_in_numbers

words_in_numbers.rb

@ones           =  {1 => "one", 2 => "two", 3 => "three", 4 => "four", 5 => "five",
                   6 => "six", 7 => "seven", 8 => "eight", 9 => "nine", 10 => "ten",
                   11 => "eleven", 12 => "twelve", 13 => "thirteen", 14 => "fourteen",
                   15 => "fifteen", 16 => "sixteen", 17 => "seventeen", 18 => "eighteen", 19 => "nineteen"}

@tens           =  {10 => "ten", 20 => "twenty", 30 => "thirty",40 => "forty", 50 => "fifty",
                   60 => "sixty", 70 => "seventy", 80 => "eighty", 90 => "ninety"}

@hundreds       =  {1 => "one hundred", 2 => "two hundred", 3 => "three hundred",
                    4 => "four hundred", 5 => "five hundred", 6 => "six hundred",
                    7 => "seven hundred", 8 => "eight hundred", 9 => "nine hundred"}



def in_words(num)
  result = []
  if num == 0
    return ""
  elsif num < 20
    result << @ones.fetch(num)
  elsif num < 100 && num > 19
    factored = num.divmod(10)
    result << @tens.fetch(factored[0] * 10) + " #{in_words(factored[1])}"
  elsif num < 1000 && num > 99
    factored = num.divmod(100)
    result << @hundreds.fetch(factored[0]) + " #{in_words(factored[1])}"
  elsif num < 1_000_000 && num > 999
    factored = num.divmod(1000)
    result << in_words(factored[0]) + " thousand" + " #{in_words(factored[1])}"
  elsif num < 1_000_000_000 && num > 999_999
    factored = num.divmod(1_000_000)
    result << in_words(factored[0]) + " million" + " #{in_words(factored[1])}"
  elsif num < 1_000_000_000_000 && num > 999_999_999
    factored = num.divmod(1_000_000_000)
    result << in_words(factored[0]) + " billion" + " #{in_words(factored[1])}"
  elsif num == 1_000_000_000_000
    result << "one trillion"
  end

  result.join
end