Using until loop in ruby problem

Question

Hey guys :) ! So I'm practicing my ruby and I'm trying to make function which sums all of the digits in a number until the sum is one digit number. Etc: 123 -> 1+2+3 = 6 99 -> 9+9 = 18 -> 1+8 =9 942 -> 9 + 4 + 2 = 15 -> 1+5 = 6 ... I'm trying to do this all using until loop but I always get first summation, my code doesn't keep to sum until i reaches one digit number. CODE:

n = gets.chomp.to_i
def digital_root(n)
#Takes n and splits its numbers into array, then it counts array
   until n.to_s.chars.map(&:to_i).count == 1
#Sums ALL the values in the previously created array
     n.to_s.chars.map(&:to_i).inject(0, :+)
     return n
   end
end

Thanks for any help :)

Answer 1 · 2016-08-10T12:39:03Z

August 10, 2016 12:39pm

Here's a solution i wrote that doesn't use the until loop...

def digital_root(n)
    number = n
    sum = 0
    #splits our string into seperate charcters then turns them back into numbers
    seperated_array = number.split(//)
    seperated_array.map! { |n| n.to_i  }
    #loop to perform our maths
    seperated_array.each do |number|
        sum += number
    end
    return sum
end
print "Enter a number: "
n = gets.chomp.to_s
puts digital_root(n)

Ill try and get my head around using the until loop to do the same thing and see if i can post a solution!

Answer 2 · 2016-08-10T12:47:41Z

August 10, 2016 12:47pm

Heres the same thing but using the until loop.

def digital_root(n)
    number = n
    sum = 0
    #splits our string into seperate charcters then turns them back into numbers
    seperated_array = number.split(//)
    seperated_array.map! { |n| n.to_i  }
    #loop to perform our maths
    i = seperated_array.size
    until i == 0
        seperated_array.each do |number|
            sum += number
             i -= 1
        end
    end
    return sum
end
print "Enter a number: "
n = gets.chomp.to_s
puts digital_root(n)

Welcome to the Treehouse Community

Looking to learn something new?

Antony Gabagul

Antony Gabagul

Using until loop in ruby problem

2 Answers

Ken jones

Ken jones

Ken jones

Ken jones