[Question]
  [
*The shortest code to pass all possibilities wins.*
In mathematics, the [persistence of a number](http://en.wikipedia.org/wiki/Persistence_of_a_number) measures how many times a certain operation must be applied to its digits until some certain fixed condition is reached. You can determine the additive persistence of a positive integer by adding the digits of the integer and repeating. You would keep adding the digits of the sum until a single digit number is found. The number of repetitions it took to reach that single digit number is the additive persistence of that number.
**Example using 84523:**
```
84523
8 + 4 + 5 + 2 + 3 = 22
2 + 2 = 4
It took two repetitions to find the single digit number.
So the additive persistence of 84523 is 2.
```
You will be given a [sequence](http://en.wikipedia.org/wiki/Sequence) of positive integers that you have to calculate the additive persistence of. Each line will contain a different integer to process. Input may be in any [standard I/O methods](https://codegolf.meta.stackexchange.com/q/2447/62402).
For each integer, you must output the integer, followed by a single space, followed by its additive persistence. Each integer processed must be on its own line.
## Test Cases
---
**Input Output**
```
99999999999 3
10 1
8 0
19999999999999999999999 4
6234 2
74621 2
39 2
2677889 3
0 0
```
  
[Answer]
## K - 29 Chars
Input is a filename passed as an argument, 29 chars not including filename.
```
`0:{5:x,-1+#(+/10_vs)\x}'.:'0:"file"
```
* 35 -> 31: Remove outside function.
* 31 -> 29: Remove parens.
[Answer]
**Python 84 Chars**
```
while 1:
 m=n=int(raw_input());c=0
 while n>9:c+=1;n=sum(map(int,str(n)))
 print m,c
```
[Answer]
## Haskell, 100 characters
```
p[d]=0
p d=1+(p.show.sum$map((-48+).fromEnum)d)
f n=n++' ':shows(p n)"\n"
main=interact$(f=<<).lines
```
[Answer]
## Python (93 bytes)
```
f=lambda n,c:n>9and f(sum(map(int,str(n))),c+1)or c
while 1:n=int(raw_input());print n,f(n,0)
```
[Answer]
# [Husk](https://github.com/barbuz/Husk), ~~10~~ 15 bytes
+5 bytes for horrible I/O requirement
```
m(wΓ·,LU¡oΣdr)¶
```
[Try it online!](https://tio.run/##yygtzv7/P1ej/NzkQ9t1fEIPLcw/tzilSPPQtv///1siAJehAZcFl6ElVsBlZmRswmVuYmZkyGVsyWVkZm5uYWHJZQAA "Husk – Try It Online")
### Explanation
To support multiple inputs, we need to use `m(₁r)¶` (where `₁` is the function doing the interesting computation):
```
m(₁r)¶  -- expects newline-separated inputs: "x₁␤x₂␤…␤xₙ"
     ¶  -- split on newlines: ["x₁","x₂",…,"xₙ"]
m(  )   -- map over each string
 ( r)   -- | read integer: [x₁,x₂,…,xₙ]
 (₁ )   -- | apply the function described below
```
The function `₁` does the following:
```
wΓ·,LU¡(Σd)  -- input is an integer, eg: 1234
      ¡(  )  -- iterate the following forever and collect results in list:
       ( d)  -- | digits: [1,2,3,4]
       (Σ )  -- | sum: 10
             -- : [1234,10,1,1,1,…
     U       -- keep longest prefix until repetition: [1234,10,1]
 Γ           -- pattern match (x = first element (1234), xs = tail ([10,1])) with:
  · L        -- | length of xs: 2
   ,         -- | construct tuple: (1234,2)
w            -- join with space: "1234 2"
```
[Answer]
### bash, 105 chars
```
while read x
do
for((i=0,z=x;x>9;i++))do
for((y=0;x>0;y+=x%10,x/=10))do :
done
x=$y
done
echo $z $i
done
```
Hardly any golfing actually involved, but I can't see how to improve it.
[Answer]
## Haskell - 114
```
s t n|n>9=s(t+1)$sum$map(read.(:[]))$show n|1>0=show t
f n=show n++" "++s 0n++"\n"
main=interact$(f.read=<<).lines
```
[Answer]
## Ruby, 85 Chars
```
puts $<.map{|n|v=n.chop!;c=0;[c+=1,n="#{n.sum-n.size*48}"] while n[1];[v,c]*' '}*"\n"
```
I had to borrow the "sum-size\*48" idea from Alex, because it's just too neat to miss (in Ruby at least).
[Answer]
## Golfscript, 40 chars
```
n%{.:${;${48-}%{+}*`:$,}%.,1>\1?+' '\n}%
```
[Answer]
## J - 45 Chars
Reads from stdin
```
(,' ',[:":@<:@#+/&.:("."0)^:a:)&><;._2(1!:1)3
```
[Answer]
## c -- 519
**(or 137 if you credit me for the framework...)**
Rather than solving just this one operation, I decided to produce a framework for solving **all persistence problems**.
```
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef char*(*O)(char*);
char*b(char*s){long long int v=0,i,l=0;char*t=0;l=strlen(s);t=malloc(l+2);
for(i=0;i<l;i++)v+=s[i]-'0';snprintf(t,l+2,"%lld",v);return t;}
int a(char**s,O o){int r;char*n;n=o(*s);r=!strcmp(*s,n);free(*s);
*s=n;return r;}
int main(int c, char**v){size_t l, m=0;char *d,*n=0;O o=b;FILE*f=stdin;
while(((l=getline(&n,&m,f))>1)&&!feof(f)){int i=0;n=strsep(&n,"\n");
d=strdup(n);while(!a(&n,o))i++;printf("%s %d\n",d,i);free(d);free(n);n=0;m=0;}}
```
**Only the two lines starting from `char*b` are unique to this problem.**
It treats the input as strings, meaning that leading "0"s are not strip before the output stage.
The above has had comments, error checking and reporting, and file reading (input must come from the standard input) striped out of: 
```
/* persistence.c
 *
 * A general framework for finding the "persistence" of input strings
 * on opperations.
 *
 * Persistence is defined as the number of times we must apply
 *
 *    value_n+1 <-- Opperation(value_n)
 *
 * before we first reach a fixed point.
 */
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "../getline.h"
/* A function pointer type for operations */
typedef char*(op_func)(char*);
typedef op_func* op_ptr;
/* Op functions must
 * + Accept the signature above
 * + return a point to a newly allocated buffer containing the updated str
 */
char* addop(char*s){
  int i,l=0;
  long long int v=0;
  char *t=NULL;
  /* protect against bad input */
  if (NULL==s) return s;
  /* allocate the new buffer */
  l = strlen(s);
  t = malloc(l+2);
  if (NULL==t) return t;
  /* walk the characters of the original adding as we go */
  for (i=0; i<l; i++) v += s[i]-'0';
  //fprintf(stderr,"   '%s' (%d) yields %lld\n",s,l,v);
  snprintf(t,l+2,"%lld",v);
  //fprintf(stderr,"   %lld is converted to '%s'\n",v,t);
  return t;
}
/* Apply op(str), return true if the argument is a fixed point fo
 * falsse otherwise,
 */ 
int apply(char**str, op_ptr op){ 
  int r;
  char*nstr;
  /* protect against bad input */
  if ( NULL==op ) exit(1); 
  if ( NULL==*str ) exit(4); 
  /* apply */
  nstr = op(*str); 
  /* test for bad output */
  if ( NULL==nstr ) exit(2); 
  r = !strcmp(*str,nstr); 
  /* free previous buffer, and reasign the new one */
  free(*str); 
  *str = nstr; 
  return r; 
}
int main(int argc, char**argv){
  size_t len, llen=0;
  char *c,*line=NULL;
  op_ptr op=addop;
  FILE *f=stdin;
  if (argc > 1) f = fopen(argv[1],"r");
  while( ((len=getline(&line,&llen,f))>1) && line!=NULL && !feof(f) ){
    int i=0;
    line=strsep(&line,"\n"); // Strip the ending newline
    /* keep a copy for later */
    c = strdup(line);
    /* count necessary applications */
    while(!apply(&line,op)) i++;
    printf("%s %d\n",c,i);
    /* memory management */
    free(c);
    free(line);
    line=NULL;
    llen=0;
  }
}
```
A little more could be saved if we were willing to leak memory like a sieve. Likewise by `#define`ing return and the like, but at this point I don't care to make it any uglier.
[Answer]
# [APL (Dyalog Unicode)](https://www.dyalog.com/), ~~21~~ 17 bytes
```
{⍵≤9:0⋄1+∇+/⍎¨⍕⍵}
```
-4 bytes from Jo King.
[Try it online!](https://tio.run/##SyzI0U2pTMzJT///qHeuJQIoGHMBBQwNFAxBtIWCAZhriRUomIAkzYyMTRSMQCxzEzMjQwjT2BJCG5mZm1tYQE01AJqm8ahvqlsQkAgI0HzUNsHQyMJcwdjkf/Wj3q2POpdYWhk86m4x1H7U0a6t/6i379CKR71TgVK1/9OAioECj7qaH/WuedS75dB640dtE4HmBAc5A8kQD8/g/2kKSI7jSlMwNAASFiAGdvcDZUCOB1JglwNpY5AY1M1AlgEA "APL (Dyalog Unicode) – Try It Online")
```
{s←+/⍎¨⍕⍵⋄⍵≤9:0⋄1+∇s}
```
[Try it online!](https://tio.run/##SyzI0U2pTMzJT///qHeuJQIoGHMBBQwNFAxBtIWCAZhriRUomIAkzYyMTRSMQCxzEzMjQwjT2BJCG5mZm1tYQE01AJqm8ahvqlsQkAgI0HzUNsHQyMJcwdjkf3UxkKOt/6i379CKR71TH/VufdTdAiI7l1haGQDZhtqPOtqLa/@nAdUBVT3qan7Uu@ZR75ZD640ftU0Emhcc5AwkQzw8g/@nKSA5kitNwdAASFiAGNj9AZQBeQJIgX0ApI1BYlC3A1kGAA "APL (Dyalog Unicode) – Try It Online")
## Explanation
```
{s←+/⍎¨⍕⍵⋄⍵≤9:0⋄1+∇s} ⍵ → input
       ⍕⍵             convert ⍵ to string
     ⍎¨               execute each character(getting digits)
   +/                 reduce to sum of digits
 s←                   assign to s
          ⍵≤9:        if number is single digit
              0       return 0, stop recursion
                1+∇s  Otherwise return 1+f(sum)
```
[Answer]
## J, 74 chars
```
i=:<;._2(1!:1)3
i&((],' ',":@(0 i.~9<[:".([:":[:+/"."0)^:(i.9)))@>@{~)i.#i
```
### Edits
* **(86 → 83)** Some Caps `[:` to Ats `@`
* **(83 → 79)** Unneeded parentheses
* **(79 → 75)** Changing `0".` to `".` simplifies things
* **(75 → 74)** Better Cutting
### E.g
```
i=:<;._2(1!:1)3
74621
39
2677889
0
i&((],' ',":@(0 i.~9<[:".([:":[:+/"."0)^:(i.9)))@>@{~)i.#i
74621 2  
39 2     
2677889 3
0 0  
```
[Answer]
# [K (ngn/k)](https://gitlab.com/n9n/k), 16 bytes
**Solution:**
```
{x,#1_(+/10\)\x} 
```
[Try it online!](https://tio.run/##y9bNS8/7/7@6QkfZMF5DW9/QIEYzpqLWyMzc3MLCUuH/fwA "K (ngn/k) – Try It Online")
**Explanation:**
```
{x,#1_(+/10\)\x} / the solution
{              } / lambda taking implicit x
      (     )\x  / iterate until convergence
         10\     / split into base-10 (123 => 1 2 3)
       +/        / sum
    1_           / drop first result (iterate returns input as first result)
   #             / count length of result
 x,              / prepend x (original input)
```
[Answer]
# [Jelly](https://github.com/DennisMitchell/jelly), 12 bytes
```
żDS$Ƭ€Ẉ’ƊK€Y
```
[Try it online!](https://tio.run/##y0rNyan8///oHpdglWNrHjWtebir41HDzGNd3kB25P///y0RQEfB0EBHwQJIWWIFOgpmRsYmOgrmJmZGhjoKxkABIzNzcwsLIMMAAA "Jelly – Try It Online")
Takes input as a list of integers. [+3 bytes](https://tio.run/##y0rNyan8///h7i1hh7Ye3eMSrHJszaOmNQ93dTxqmHmsyxvIjvz//7@SkpIlAnAZGnBZcBlaYgVcZkbGJlzmJmZGhlzGllxGZubmFhaWXAZAIwA) to input as a multiline string.
+3 bytes (or +6) for restrictive I/O formats, yay
## How it works
```
żDS$Ƭ€Ẉ’ƊK€Y - Main link. Takes a list l on the left
        Ɗ    - To l:
   $         -   Group the previous 2 links into a monad f(n):
 D           -     Digits
  S          -     Sum
    Ƭ€       -   Over each n in l, repeatedly apply f(n) until a fixed point,
                  yielding [n, f(n), f(f(n)), ...]
      Ẉ      -   Get the length of each loop
       ’     -   Decrement each
ż            - Zip the loop lengths with l
         K€  - Join each by spaces
           Y - Join by newlines
```
[Answer]
# [Vyxal](https://github.com/Vyxal/Vyxal) `aj`, ~~9~~ 8 bytes
```
ƛ⁽∑↔L‹"Ṅ
```
[Try it Online!](https://vyxal.pythonanywhere.com/#WyJhaiIsIiIsIsab4oG94oiR4oaUTOKAuVwi4bmEIiwiIiwiOTk5OTk5OTk5OTlcbjEwXG44XG4xOTk5OTk5OTk5OTk5OTk5OTk5OTk5OVxuNjIzNFxuNzQ2MjFcbjM5XG4yNjc3ODg5XG4wIl0=)
Explanation:
```
          # Treat all the inputs as a list (`a` flag)
ƛ         # On each item (n) in that list
 ⁽        # Create a single element lambda
  ∑       # That sums the digits of a number
   ↔      # Apply that function on n until it doesn't change 
          # and return the intermediate values in a list
    L     # Get the length of that list
     ‹    # Decrement it to ignore the last time the function is called
      "   # Pair that with n
       Ṅ  # Join the pair by a space
          # Join the mapped inputs by new lines (`j` flag)
          # Implicitly print
```
[Answer]
I think this is about the best I can come up with.
**Ruby 101 Chars**
```
f=->(n){n.sum-n.size*48}
$<.each{|l|i=0;i+=1 while(i+=1;n=f[(n||l.chop!).to_s])>10
puts "#{l} #{i}"}
```
[Answer]
**PARI/GP 101 Chars**
```
s(n)=r=0;while(n>0,r+=n%10;n\=10);r
f(n)=c=0;while(n>9,c++;n=s(n));c
while(n=input(),print(n," ",f(n)))
```
Unfortunately, there's no input function for GP, so i guess this lacks the IO part. :( 
Fixed: Thanks Eelvex! :)
[Answer]
**Javascript - 95**
```
i=prompt();while(i>9){i=''+i;t=0;for(j=0;j<i.length;j++)t+=parseInt(i.charAt(j));i=t;}alert(t);
```
EDIT: Whoops does'nt do the multi-lines
[Answer]
# J, 78
```
f=:[:+/"."0&":
r=:>:@$:@f`0:@.(=f)
(4(1!:2)~LF,~[:":@([,r)".@,&'x');._2(1!:1)3
```
Recursive solution. Reads from stdin. **Writes to stdout**, so cut me some slack - it does take an extra 18-ish characters.
[Answer]
## Perl - 77 characters
```
sub'_{split//,shift;@_<2?0:1+_(eval join'+',@_)}chop,print$_,$",(_$_),$/for<>
```
[Answer]
### scala 173:
```
def s(n:BigInt):BigInt=if(n<=9)n else n%10+s(n/10)
def d(n:BigInt):Int=if(n<10)0 else 1+d(s(n))
Iterator.continually(readInt).takeWhile(_>0).foreach(i=>println(i+" "+d(i)))
```
[Answer]
# [JavaScript](https://nodejs.org), ~~57~~ 47 bytes
*-10 bytes thanks to @l4m2!*
```
f=(s,c=0)=>s>9?f(eval([...s+""].join`+`),++c):c
```
[Try it online!](https://tio.run/##fdDRCoMgFAbg@z1FeKXYTC0sB7YHGYPCVRSRY45e39XVwqz/@jv/4ZyhnmurP/37e53Mq3GuVdDGWlGkSlvKewubuR7hgxBiMQBPMph@qnCFYow1ummnzWTN2JDRdLCFQP4DEIq2SZIovXicUV9tOfN5cahXTnftMpi1ZOGZzwVPs4MFC@c@zzPBWdiHeCrPTt1xLvK8KIIzoUfS88@4Hw "JavaScript (Node.js) – Try It Online")
[Answer]
# [MathGolf](https://github.com/maxbergmark/mathgolf), 11 bytes
```
hÅ_Σ]▀£(k ?
```
[Try it online!](https://tio.run/##y00syUjPz0n7/z/jcGv8ucWxj6Y1HFqska1g//@/JQJwGRpwWXAZWmIFXGZGxiZc5iZmRoZcxpZcRmbm5hYWllwG/wE "MathGolf – Try It Online")
~~Incredibly inefficient, but we don't care about that. Basically, using the fact that the additive persistence of a number is smaller than or equal to the number itself.~~
Uses the fact that the additive persistence is less than or equal to the number of digits of the number. Passes all test cases with ease now.
The input format, while suboptimal for some languages, is actually the standard method of taking multiple test cases as input in MathGolf. Each line of the input is processed as its own program execution, and output is separated by a single newline for each execution. 
## Explanation (using `n = 6234`)
```
h             push length of number without popping (6234, 4)
 Å            loop 4 times using next 2 operators
  _           duplicate TOS
   Σ          get the digit sum
    ]         wrap stack in array
              this gives the array [6234, 15, 6, 6, 6]
     ▀        unique elements of string/list ([6234, 15, 6])
      £       length of array/string with pop (3)
       (      decrement (2)
        k ?   push input, space, and rotate top 3 elements to produce output (6234 2)
```
[Answer]
# Julia, 92 (29) bytes
```
f(n)=n>9&&(f∘sum∘digits)(n)+1
```
Edit:
With correct printing it's 92:
```
f(n)=n>9&&(f∘sum∘digits)(n)+1
while(n=parse(Int128,readline()))≢π
println("$n ",f(n)%Int)end
```
[Answer]
# [Stax](https://github.com/tomtheisen/stax), ~~8~~ 11 [bytes](https://github.com/tomtheisen/stax/blob/master/docs/packed.md#packed-stax)
```
ªwæMε∞ö?îm⌐
```
[Run and debug it](https://staxlang.xyz/#p=a677914deeec943f8c6da9&i=99999999999%0A10%0A8%0A19999999999999999999999%0A6234%0A74621%0A39%0A2677889%0A0&a=1)
+3 bytes thanks to @Khuldraeseth (the first answer didn't have compliant output)
[Answer]
# [Jq](https://stedolan.github.io/jq/) `-r`, 52 bytes
```
[while(length>1;explode|map(.-48)|add|@text)]|length
-r                    # take one number per line, converting it to a string
                explode                              # the character codes
                       |map(.-48)                    # subtract ascii zero from each
                                 |add                # sum the elements
                                     |@text          # back to text, (shorter than
                                                     # tostring and tojson)
 while(length>1;                           )         # produces *all* the intermediate
                                                     # results while the string length>1
[                                           ]        # collected into an array
                                             |length # the length of these results
```
Not answering the multiplicative persistence now because jq doesn't have a `product` builtin :)
[Jq play it!](https://jqplay.org/s/HjHvWmd9d7)
[Answer]
# [05AB1E](https://github.com/Adriandmen/05AB1E/wiki/Commands), ~~13~~ 10 [bytes](https://github.com/Adriandmen/05AB1E/wiki/Codepage)
```
ε.ΓSO}g}ø»
```
-3 bytes thanks to *@Steffan*
Input as a list of integers.
[Try it online.](https://tio.run/##yy9OTMpM/f//3Fa9c5OD/WvTaw/vOLT7//9oSwTQMTTQsdAxtMQKdMyMjE10zE3MjAx1jC11jMzMzS0sLHWAOkxMjYxjAQ)
**Explanation:**
```
ε      # Map each integer in the (implicit) input to:
 .Γ    #  Loop until the integer no longer changes,
       #  keeping all intermediate steps in a list:
   S   #   Convert the integer to a list of digits
    O  #   Sum those digits together
  }g   #  After the inner loop: pop and push the length
}ø     # After the outer map: zip/transpose it with the (implicit) input to create a
       # list of pairs
  »    # Join each pair by a space, and then the list of strings by newlines
       # (after which the result is output implicitly)
```
[Answer]
# [Knight](https://github.com/knight-lang/knight-lang), 54 bytes
```
;=iP;=r 0;W<1Li;=s 0;Wi;=s+sA Ai=iGi 1Li;=i+""s=r+1rOr
```
[Try it online!](https://tio.run/##rVltV9vKEf5s/4rFOdhavySSoOfcWsgckgscTgnkQtJ8ADcRsgx7KsuuJMOlhP71dGb2RSvbpE17@SBpd@f12dmZWRMPbuP4@yuRxelykuwV5UTMX9@NmvZMKm5qU3H5uEhWiHKR3damSjGTNJNkKrKEpSlL59ktPZpm9uPZp/fMq4aXHy@YXw3/enDBdqvh0dk79ks1vPjEvIr46PTSoj37dGqRfjp7f3D5F6fgzMFHm/3L@xM3qweXoFYuxndRzrq8YrCpwFYjYjRiu9Xa2eFnXMxgkTz6xvB7b2@FBtUQDboJNGnKkbBG8/b8/JSI8LFPTg7RN8uQi2NJAPyWrfdRukw4v8rGvNkkP2BXrsbhRevw/Mj5HoTiQxDmzA0@73mnIggL/MR3rzhgByIUx4LRgui1WkWY97z8PP/OgbkVSHldEJhEMxai5KDZBPWLKC8Sh7OnZkNkJRNBswGzZEqfdafLLIYZYo5hXM4W/WKRXfnj8Mntu88goyFAHNHD28WJN29QvFiwhztRJsUiipNmA77TxIF5YHekGX3Wui6vs@vpdc6enq/GDh@@anEypSGmzDHWhqzzqsOZFKFnt2D2OoPpXk/OgJ2NJC0Se@JZmhPfJfHf2XSes2w5u0nyQtnDHFFMxK0otVTQju54feORfHeZ57Ie6yrLez3OBqzjdkAFWio4y5NymWdMB49ko/gJVm24j3IR3aSJtgKMSOcPSe7EoE8bwr59Y9q4mEYxAfGlwxU8YoPrIvT6sEehmV6xDo@jClsgySbLhYNbyhSmAwajdYNlciikNGnGdadT2dTqyE0DZvDAGMVACCQT@AbLFlFRsvIuAWFRXgKx2oCuRhQ3NOboiTbWOmMvGgtPj9NGNyowyHwM3VLMMzAbI9Ydg2mxRqRYLhYIONdRpWds@O34A9gtsFGGDuVFBnEMiQ0jV1kuUfnYMadfThzhBCa5ISY3NFMHGeVF6aqDdqOpUZrOY2fX7XscHcRp4wQ5mC3TNMofbUer/bmwtudDx1hGChX/MlvhJhUeqlBJYZOnB4dv3339bev013PLYVvsjdgo11@Xu1UTfHxy@YLEMslJZJRN2D@WkR6i2Htbxc66CgnAccdgcbIBC2LerTPXaZ4pVebLDHcoUMm5W84LR@dKOgAQ2qWIGa3eLKdX/u5Y2SE3uk3pwRhQLOBYlVMHSMH/7TSdtHTeoeLURyGrAuBQGAGy7En9Nbo95u3U/UTnMRb3WavMl0kLYtDMY0jC/DSCDIILLYysVgUC@om@q/pEcHx5O5@nsHJTR@BFXyu3fuRQ9@c9ss28WTMzRRuzP9RGCNpynqaObWqfuX2oEP@DydmayViF3x98OOtjLZaBBsOTK891XQgncAWGelSIfyZfSoavwIpRy1tEAEZ9fPj03Ak9XdIxw2FVxxc9ffnaAYKjk9PDLptCcgwaJrSVPkzDcbTop0lGHYCFlW6fJGZw7l4CY5URaxOWNiw4jqBeAmrxHmERQFERuGjyRjxbrOwAgSTGVQpBmMQY1RQPoozvnC4gs9ZtAUZM/cVYpzoHnSFDLZIgRECxX4MszNvklBKvOsZayPKAqnBFQW1pmWPpos70asyfqjCDBopyu1R8AYrrsnPIeA5vu79P8a@i/ACUIFVidJuUKTSVTpv2so07hCVyIjIeiKnj0H5iQxDf5WhLH@KUc7nNoRtstBVenNoAqe9Q6pOdI2YDBUig7UWQVO6srHxb@SPpzcq7agU5bYgNzVegwWiA@AsX80WSOZbifitvUasAVoUzWSkhHkPf3f2F5lVz4ciOZAqmT9ApaOEgZPvY3wE59A40Qh2gWoeXA5OsF1IrhCcWKLmCG@SQLoIaJXRD5ts9SwVj5clv4EnyO7RxeNrX/Nyq7TrdHbYwk60Rnq6FB2hJ67hY5L92hqpP3BDIJvOpCtQ6o7bYwRrEdREajXZ51ViupUPNeknNobNdIGP9MKwzA5zQ@th6IZPyVkWqF7C8JFGmxL5UwlTNsndA5RaFwnmFAh4MefFB0OR@1qBj7TZduAYDJB3LNvea2u2GNmv7dbcAexwYcaaPWm6ZvywLvfsmJmSvJw3qGYMoT0K@xJSMoWVtj@ymG/XdlrdXpIa@n@lY8uWeY/9Lh0FuAFHhNJ7yUHvp882hGkelfqmu0@tVMHH9jbI4dKN9epcqRyi/BmvRaUU7tup1eyVTt9qdDSislucNOHTXcJD7ugmF2QbXukQT2kKIAWqQQxky7FzDJQ@JgsGA5CmgqGSubrREtKyd/jc/BObNZmC2f8i0vZnpb4RmzSF5daA53/YyULBL1EM28AzKRAvQe3DOBh6cMc8m9n@Seg/SY402DJGU9A71jKshJ6D9kauw9jm1K92Q8G@uHghaJFfgjkAJFksMfOOV4S7Jk07Bsjn00CItRcYeokfAjU3mcCctk9skB57FPEuyUkR4i4CTPGcPCbuL7hMgVIKQvGSzKFtC@Dy@xln07gYRBygIiWKZlvijASk3CU@TDGwarXIFMpvPkKyIR3kvU7ovG2LI9mpULt3PoxRr4yN4nU3SZMK@VmZ/NVKe8Es2ZpaSypkRChsM9JgTeUORdi3Tn00e3KNgXSl8mPRqeTGsZ8V9Zh/@PfsYIOtQ/S7XZpJgn9UaRZrsMysbogxXcuqS68s6gLRbIVQbqwsawaHE0JpP5kM2S2bxAsMsY6Ov@/@vK6M/wJWRdmVrgy@h8kW5sv8y@lLrBpNxeZ/VCHxD4esCurXJzmpg3NOZxxZgzGtX6Q992VQb9m1OlUoM/7f/hl/lIEuM4Q86Q5u61uau0oZV1q0v//RFRve21U0G6Rq124xJhUSDEsdhjdtKk8iBFL2eYZMmfwb35K9edqNZ847ZvYtiO6l37Uhq8e/7wx0LlmN5ZQitDqtnF6D1O1KmLh59cxZ2ag3tpQR6RWiwWvL8TSVvh5s@wDfcuzT5HxoiNAzfuBWxaf5NO2T3Rj54iFp7HjZJ1CDI@JeL9CKKnjoJoH2STCNIkuCZ7jBFBotiQj8iRXEJZaqzHXew58QZzl64xaIwUwOhG1Q/JMwikWGzyqL8Nu6T0G4Xvu85/lxF90v8N4/jkjX1a9zzd@/PG//@DQ)
[Answer]
# [Ly](https://github.com/LyricLy/Ly), 21 bytes
```
0s+1[pSy,![lu;]&+l`s]
```
[Try it online!](https://tio.run/##y6n8/9@gWNswuiC4UkcxOqfUOlZNOyehOBYoDAA "Ly – Try It Online")
A pretty straight forward interpretation of the challenge rules this time.
```
0s+1[pSy,![lu;]ç]
0s                    - initialize the backup cell to 0 (iteration count)
  +                   - load input number, absorbing the 0 on the stack
   1                  - push a truthy value to enter loop
    [p              ] - infinite loop
      S               - convert number on top of stack to digits
       y              - push the number of digit (stack size)
        `!            - decrement and convert truthiness
          [   ]       - if/then, executes is number is a single digit
           lu;        - load backup cell, print as number, end program
               &+     - sum the digits on the stack
                 l`s  - load backup cell (iterations) increment and save
```
]