Solution: Cryptic Math Equation (Backtracking)

@letters = ['U', 'K', 'S', 'A', 'R', 'G', 'I', 'N']
@letter_to_digits = {} # hash to store the solution
@is_digit_used = {0 => false, 1 => false, 2 => false, 3 => false, 4 => false, 5 => false, 6 => false, 7 => false, 8 => false, 9 => false }
def decode_letters_to_number(a, d)
  d.each do |k, v| 
    a = a.gsub(k,v)   #if k && v
  end  
  a.to_i
end
def check_ans()
  a = decode_letters_to_number('UK', @letter_to_digits)
  b = decode_letters_to_number('USA', @letter_to_digits)
  c = decode_letters_to_number('USSR', @letter_to_digits)
  d = decode_letters_to_number('AGING', @letter_to_digits)
  # puts "[DEBUG] Checking #{a} + #{b} + #{c} = #{d}"
  if a + b + c  == d
    puts "#{a} + #{b} + #{c} = #{d}" # got it
    puts @letter_to_digits
  end
end
def find_out(idx = 0)
  if idx == @letters.length # got 8 letter filled
    check_ans()
    return
  end
  (0..9).each do |i|  
    # assign 0 - 9 to each letter
    next if i == 0 and (@letters[idx] =='U' || @letters[idx] =='A' ) # no leading 0
    next if @letters[idx] == 'A' && i != 1
    next if @letters[idx] == 'U' && i != 9
    next if @letters[idx] == 'G' && i != 0
    
    unless @is_digit_used[i]
      @is_digit_used[i] = true  # the number is used
      @letter_to_digits[@letters[idx]] = i.to_s
      find_out(idx + 1)
      @is_digit_used[i] = false # clear
    end
  end
end
start_time = Time.now
find_out # start
puts "Time to solve: #{Time.now - start_time}"