Print the digital root

Question

^{This is different from My Word can beat up your Word as it is less complex and only requires you to calculate it, and not compare them.}

To find the digital root, take all of the digits of a number, add them, and repeat until you get a one-digit number. For example, if the number was 12345, you would add 1, 2, 3, 4, and 5, getting 15. You would then add 1 and 5, giving you 6.

Your task

Given an integer N (0 <= N <= 10000) through STDIN, print the digital root of N.

Test cases

1 -> 1
45 -> 9
341 -> 8
6801 -> 6
59613 -> 6
495106 -> 7

Remember, this is code-golf, so the code with the smallest number of bytes wins.

Answer 1

Pyke, 1 byte

s

Try it here!

Takes the digital root of the input

Answer 2

Jelly, 7 5 4 3 bytes

ḃ9Ṫ

TryItOnline! or all test cases

How?

The digital root is known to obey the formula (n-1)%9+1.
This is the same as the last digit in bijective base 9
(and due to implementation that 0ḃ9=[] and []Ṫ=0 this handles the edge-case of zero).

ḃ9Ṫ - Main link: n
ḃ9  - convert to bijective base 9 digits (a list)
  Ṫ - tail (get the last digit)

Answer 3

JavaScript (ES6), 16 10 bytes

n=>--n%9+1

Test cases

let f =

n=>--n%9+1

console.log(f(1)); // -> 1
console.log(f(45)); // -> 9
console.log(f(341)); // -> 8
console.log(f(6801)); // -> 6
console.log(f(59613)); // -> 6
console.log(f(495106)); // -> 7

Answer 4

Python, 16 20 bytes

+4 bytes to handle edge case of zero.

lambda n:n and~-n%9+1

repl.it

Answer 5

MATL, 3 bytes

9X\

Try it online!

A lot of (now deleted answers) tried using modulo 9 to get the result. This is a great shortcut, but unfortunately does not work for multiples of 9. MATL has a function for modulo on the interval [1, n]. Using this modulo, we have 1 % 3 == 1, 2 % 3 == 2, 3 % 3 == 3, 4 % 3 == 1, etc. This answer simply takes the input modulo nine using this custom modulo.

Answer 6

Mathematica, 27 11 bytes

Mod[#,9,1]&

Mathematica's Mod takes a third parameter as an offset of the resulting range of the modulo. This avoids decrementing the input and incrementing the output.

Answer 7

Julia, 12 bytes

!n=mod1(n,9)

or

n->mod1(n,9)

mod1 is an alternative to mod which maps to the range [1, n] instead of [0, n).

Answer 8

PHP, 15 Bytes

<?=--$argn%9+1;

Previous version PHP, 55 Bytes

$n=$argn;while($n>9)$n=array_sum(Str_split($n));echo$n;

Answer 9

R, 72 67 29 bytes

Edit: Thanks to @rturnbull for shaving off two bytes.

n=scan();`if`(n%%9|!n,n%%9,9)

Answer 10

Retina, 7 bytes

{`.
*
.

Try it online!

I see lots of mathematical solutions, but in Retina the straightforward approach seems to be the best one.

Explanation

{` makes the whole program run in a loop until the string doesn't change anymore. The loop consists of two stages:

.
*

Convert each digit to unary.

.

Count the number of characters (=convert the unary number to decimal).

This works because converting each digit to unary with no separator between digits creates a single unary number which is equal to the sum of all digits.

Answer 11

Haskell, 35 34 bytes

until(<10)$sum.map(read.pure).show

Try it on Ideone.

Explanation:

until(<10)$sum.map(read.pure).show
                              show  -- convert int to string
               map(         ).      -- turn each char (digit) into
                        pure        --    a string 
                   read.            --    and then a number
           sum.                     -- sum up the list of numbers
until(<10)$                         -- repeat until the result is < 10

Answer 12

Perl, 15 bytes

Includes +2 for -lp

Give input on STDIN

root.pl <<< 123

root.pl

#!/usr/bin/perl -lp
$_&&=~-$_%9+1

This is the boring solution that has already been given in many languages, but at least this version supports 0 too

More interesting doing real repeated additions (though in another order) is in fact only 1 byte longer:

#!/usr/bin/perl -p
s%%$_+=chop%reg

Answer 13

Husk, 4 bytes

Recursive method (4 bytes) (Credit to @ovs)

ωoΣd

      # implicit parameter ⁰ (refers to the last parameter)
   d  # list of digits in base 10
  Σd  # sum of list
 oΣd  # compose the two functions
ωoΣd  # repeat until the function returns the same result as the previous one

Try it online!

Modulus method (4 bytes)

→%9←

      # implicit parameter ⁰ (refers to the last parameter)
   ←  # previous number (n - 1)
 %9←  # modulus by 9
→%9←  # next number (n + 1)

Try it online!

Answer 14

Java 7, 63 bytes

int f(int n){int s=0;for(;n>0;n/=10)s+=n%10;return s>9?f(s):s;}

Recursive function which just gets digits with mod/div. Nothing fancy.

Cheap port

of Jonathan Allan's would be a measly 28 bytes:

int f(int n){return~-n%9+1;}

Answer 15

Labyrinth, 8 bytes

?(_9%)!@

using the equation (n-1)%9+1:

• ? reads the input as decimal and pushes it to the stack
• ( decrements the top of the stack
• _ pushes a zero onto the top of the stack
• 9 push the top of the stack popped times 10 the digit (in this case, 9)
• % pops y, pops x, pushes x%y
• ) increments the top of the stack
• ! pops the top of the stack and out puts it as a decimal string
• @ terminates the program

Answer 16

Hexagony, 19 15 bytes

.?<9{(/>!@!/)%' 

More Readable:

  . ? < 
 9 { ( /
> ! @ ! / 
 ) % ' .
  . . . 

Try it online!

-3 bytes by taking a different approach, making the 0 edge case trivial.
-1 byte by fixing 0 edge case bug

Using the formula ((n-1) mod 9) + 1 like a lot of other solutions aswell.

Answer 17

Japt -h, 6 4 bytes

Æ=ìx

Try it

Æ=ìx     :Implicit input of integer U
Æ        :Map the range [0,U)
 =       :  Reassign to U
  ì      :  Convert to digit array
   x     :  Reduce by addition
         :Implicit output of last element

Answer 18

Brachylog, 9 bytes

#0|@e+:0&

Try it online!

Explanation

#0            Input = Output = a digit
  |           OR
   @e         Split the input into a list of digits
     +        Sum
      :0&     Call this predicate recursively

Alternative approach, 11 bytes

:I:{@e+}i#0

This one uses the meta-predicate i - Iterate to call I times the predicate {@e+} on the input. This will try values of I from 0 to infinity until one makes it so that the output of i is a single digit which makes #0 true.

Answer 19

05AB1E, 6 bytes

[SODg#

Try it online!

Explanation

[        # infinite loop
 S       # split into digits
  O      # sum digits
   Dg#   # if length == 1: break

Answer 20

Ruby, 12 bytes

->n{~-n%9+1}

Answer 21

JavaScript (ES6), 41 38 bytes

Saved 3 bytes, thanks to Bassdrop Cumberwubwubwub

Takes and returns a string.

f=s=>s[1]?f(''+eval([...s].join`+`)):s

Test cases

f=s=>s[1]?f(''+eval([...s].join`+`)):s

console.log(f("1")); // -> 1
console.log(f("45")); // -> 9
console.log(f("341")); // -> 8
console.log(f("6801")); // -> 6
console.log(f("59613")); // -> 6
console.log(f("495106")); // -> 7

Answer 22

CJam, 19 13 bytes

r{:~:+_s\9>}g

Interpreter

Explanation:

r{:~:+_s\9>}g Code
r             Get token
 {:~:+_s\9>}  Block: :~:+_s\9>
   ~          Eval
  :           Map
     +        Add
    :         Map
      _       Duplicate
       s      Convert to string
        \     Swap
         9    9
          >   Greater than
            g Do while (pop)

Thanks to 8478 (Martin Ender) for -6 bytes.

---

CJam, 6 bytes

ri(9%)

Suggested by 8478 (Martin Ender). Interpreter

I was thinking about it, but Martin just got it before me. Explanation:

ri(9%) Code
r      Get token
 i     Convert to integer
  (    Decrement
   9   9
    %  Modulo
     ) Increment

Answer 23

Factor, 24

Smart, mathy answer.

[ neg bitnot 9 mod 1 + ]

63 for dumb iterative solution:

[ [ dup 9 > ] [ number>string >array [ 48 - ] map sum ] while ]

Answer 24

J, 8 bytes

**1+9|<:

Uses the formula d(n) = ((n-1) mod 9) + 1 with the case d(0) = 0.

Usage

   f =: **1+9|<:
   (,.f"0) 0 1 45 341 6801 59613 495106
     0 0
     1 1
    45 9
   341 8
  6801 6
 59613 6
495106 7

Explanation

**1+9|<:  Input: integer n
      <:  Decrement n
    9|    Take it modulo 9
  1+      Add 1 to it
*         Sign(n) = 0 if n = 0, else 1
 *        Multiply and return

Answer 25

Pyth - 7 4 6 7 bytes

Not the best one, but still beats a decent amount of answers:

|ejQ9 9

Like the previous version, but handling also cases of multiples of 9, using logical or.

---

This version fails the 45 testcase:

ejQ9

Explanation:

 jQ9  -> converting the input to base 9
e     -> taking the last digit

Try it here

Try the previous version here!

---

Previous solutions:

&Qh%tQ9

Explanation:

    tQ    -> tail: Q-1
   %tQ9   -> Modulo: (Q-1)%9
  h%tQ9   -> head: (Q-1)%9+1
&Qh%tQ9   -> Logical 'and' - takes the first null value. If Q is 0 - returns zero, otherwise returns the (Q-1)%9+1 expression result

You're invited to try it here!

Answer 26

APL (Dyalog), 15 9 bytes bytes

××1+9|-∘1

Try it online!

Answer 27

Python 2, 54 51 bytes

i=input()
while~-len(i):i=`sum(map(int,i))`
print i 

Thanks to Oliver and Karl Napf for helping me save 3 bytes

Answer 28

Python, 45 bytes

f=lambda x:x[1:]and f(`sum(map(int,x))`)or x

Takes the argument as a string.

Answer 29

C, 64 29 bytes

C port from Jonathan Allan's answer (with special case 0).

f(i){return i>0?~-i%9+1:0;}

Previous 64 byte code:

q(i){return i>9?i%10+q(i/10):i;}
f(i){i=q(i);return i>9?f(i):i;}

q takes the cross sum and f repeats taking the cross sum until a single digit.

Answer 30

Retina, 15 bytes

.+
$*
1{9}\B

1

Try it online! (The first line enables a linefeed-separated test suite.)

Explanation

.+
$*

Convert input to unary.

(1{9})*\B

Take 1-based modulo by removing nines that have at least one more character after them.

1

Count the remaining number of 1s to convert back to decimal.