Persistance.rb

=begin

Write a function, persistence, that takes in a positive parameter num and
returns its multiplicative persistence, which is the number of times you must
multiply the digits in num until you reach a single digit.

For example:

persistence(39) # returns 3, because 3*9=27, 2*7=14, 1*4=4
                 # and 4 has only one digit

persistence(999) # returns 4, because 9*9*9=729, 7*2*9=126,
                  # 1*2*6=12, and finally 1*2=2

persistence(4) # returns 0, because 4 is already a one-digit number

=end

# My Solution
def persistence(n, amount=0)
    return amount if n < 10

    n = n.to_s.split("")
    num = n[0].to_i

    1.upto(n.length-1) {|x| num = num * n[x].to_i}

    amount += 1
    persistence(num,amount)
end

# Better Solution
def persistence(n)
  n < 10 ? 0 : 1 + persistence(n.to_s.chars.map(&:to_i).reduce(:*))
end

# Another Solution
def persistence(num)
  if num < 10 then
    return 0
  else
    return 1 + persistence(num.to_s.chars.inject(1) { |n,c| n * c.to_i })
  end
end