Given an integer n > 9, for each possible insertion between digits in that integer, insert an addition + and evaluate. Then, take the original number modulo those results. Output the sum total of these operations.

An example with n = 47852:

47852 % (4785+2) = 4769
47852 % (478+52) =  152
47852 % (47+852) =  205
47852 % (4+7852) =  716


A single positive integer in any convenient format, n > 9.


The single integer output following the above construction technique.


  • You don't need to worry about input larger than your language's default type.
  • Either a full program or a function are acceptable. If a function, you can return the output rather than printing it.
  • Standard loopholes are forbidden.
  • This is so all usual golfing rules apply, and the shortest code (in bytes) wins.


47852 -> 5842
13 -> 1
111 -> 6
12345 -> 2097
54321 -> 8331
3729105472 -> 505598476

37 Answers 37


05AB1E, 12 10 bytes


Uses the CP-1252 encoding. Try it online!

  • 3
    \$\begingroup\$ nice, didn't reload and posted D.s¨s.p¨R+¹s%O without seeing this ;P \$\endgroup\$ Dec 9, 2016 at 15:39

JavaScript, 43 47 bytes

n=>eval(n.replace(/./g,'+'+n+"%($`+ +'$&$'')"))

<input id=I>
<input id=O disabled>

Takes input as string.


+4 bytes: Leading zeroes in JavaScript converts the number to octal ):

  • 2
    \$\begingroup\$ That snippet is pretty neat, seeing it update in real-time like that. \$\endgroup\$ Dec 9, 2016 at 15:05
  • \$\begingroup\$ Can you save a byte by doing (+'$&$''+$`)? \$\endgroup\$
    – Neil
    Dec 9, 2016 at 16:48
  • \$\begingroup\$ @Neil. In the first iteration $` is empty, and it will throw error trying to eval (13+) (as example). \$\endgroup\$ Dec 9, 2016 at 17:37
  • \$\begingroup\$ Can you please explain why it doesn't work only for last test case when I write $`+$&$'? \$\endgroup\$
    – EzioMercer
    Mar 2, 2023 at 9:24
  • 1
    \$\begingroup\$ @EzioMercer It is because when you do (37291+05472) the second number is converted to octal, because it starts with zero. \$\endgroup\$ Mar 2, 2023 at 12:35

Brachylog v1, 20 bytes


Try it online!


This implements the formula given. The only thing we have to be careful about is when a 0 is in the middle of the input: in that case Brachylog gets pretty quirky, for example it won't accept that a list of integers starting with a 0 can be concatenated into an integer (which would require ignoring the leading 0 — this is mainly programmed that way to avoid infinite loops). Therefore to circumvent that problem, we convert the input to a string and then convert back all splitted inputs into integers.

                       Example Input: 47852

:{               }f    Find all outputs of that predicate: [716,205,152,4769]
  $@                     Integer to String: "47852"
    ~c#C                 #C is a list of two strings which when concatenated yield the Input
                           e.g. ["47","852"]. Leave choice points for all possibilities.
        :@$a             Apply String to integer: [47,852]
            +            Sum: 899
             :?r%        Input modulo that result: 205
                   +   Sum all elements of that list               

ES6 (Javascript), 42, 40 bytes


  • Got rid of s, -2 bytes






  • \$\begingroup\$ If you limit yourself to m<2**31 then you can start with x=1 for a saving of a byte. \$\endgroup\$
    – Neil
    Dec 9, 2016 at 19:33

Jelly, 12 bytes




ŒṖṖLÐṂḌS€⁸%S - Main link: n
ŒṖ           - partitions (implicitly treats the integer n as a list of digits)
  Ṗ          - pop - remove the last partition (the one with length one)
    ÐṂ       - keep those with minimal...
   L         - length (now we have all partitions of length 2)
      Ḍ      - undecimal (convert each entry back to an integer)
       S€    - sum each (add the pairs up)
         ⁸   - left argument, n
          %  - mod (vectorises)
           S - sum

Python 2, 45 bytes

f=lambda n,c=10:n/c and n%(n/c+n%c)+f(n,c*10)

Uses arithmetic rather than strings to divide the input n into parts n/c and n%c, which c recurses through powers of 10.


Perl 35 32 27 Bytes

Includes +3 for -p

Saved 8 bytes thanks to Dada


C 77 + 4 = 81 bytes


i,N,t,r,m;f(n){for(r=0,m=n;n;t=n%10,n/=10,N+=t*pow(10,i++),r+=m%(n+N));return r;}  




    return r;

  • \$\begingroup\$ You should init r=0 such that if the function is called again the result is correct. It is somewhere in Meta, if you use global variables then you have to deal with the side effects of calling a function more than once. \$\endgroup\$
    – Karl Napf
    Dec 10, 2016 at 18:58
  • \$\begingroup\$ @KarlNapf that good? \$\endgroup\$ Dec 10, 2016 at 19:07
  • \$\begingroup\$ C does not allow default function values, your code does not compile. You can declare r global but inside the function as a statement you can say r=0;, see my answer for example. \$\endgroup\$
    – Karl Napf
    Dec 10, 2016 at 19:14
  • 1
    \$\begingroup\$ @KarlNapf your ans is v2 of my ans...far better thanks \$\endgroup\$ Dec 10, 2016 at 19:16

Python 2, 68 64 68 bytes

-4 bytes thanks to atlasologist

lambda x:sum(int(x)%(int(x[:i])+int(x[i:]))for i in range(1,len(x)))

*Input is a string

  • \$\begingroup\$ Save 4: lambda n:sum(int(n)%eval(n[:i]+'+'+n[i:])for i in range(len(n))) \$\endgroup\$ Dec 9, 2016 at 15:35
  • 1
    \$\begingroup\$ Fails for inputs containing a zero with an 8 or 9 after it and gives wrong answers for others (like the last test case). Numbers starting with a zero are octal. repl.it/EmMm \$\endgroup\$
    – mbomb007
    Dec 9, 2016 at 21:11
  • \$\begingroup\$ @mbomb007 fixed \$\endgroup\$
    – Rod
    Dec 13, 2016 at 17:53

C, 59 bytes

t,s;f(n){s=0;t=10;while(t<n)s+=n%(n/t+n%t),t*=10;return s;}

t is 10,100,1000,... and represents the cut in the big number. n/t is the right part and n%t the left part. If t is bigger than the number, it is finished.

Ungolfed and usage:

  s += n % (n/t + n%t),
  t *= 10;
 return s;

  • \$\begingroup\$ Ohhh my....please add an explanation. \$\endgroup\$ Dec 10, 2016 at 19:09
  • \$\begingroup\$ @MukulKumar okay like this? \$\endgroup\$
    – Karl Napf
    Dec 10, 2016 at 19:19
  • \$\begingroup\$ yeah that's nice. \$\endgroup\$ Dec 10, 2016 at 19:21

Retina, 38 bytes

Byte count assumes ISO 8859-1 encoding.



Not exactly efficient...

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



Between every pair of characters, we insert a comma, everything in front of the match, a semicolon, the entire input, a linefeed, and everything after the match. For input 12345 this gives us:


I.e. every possible splitting of the input along with a pair of the input. We don't need that last line though so:


We discard it.


This replaces each number as well as the comma with its unary representation. Since the comma isn't a number, it's treated as zero and simply removed. This adds the two parts in each splitting.


This computes the modulo by removing all copies of the first number from the second number.


That's it, we simply count how many 1s are left in the string and print that as the result.


Pyth, 14 bytes


A program that takes input of an integer and prints the result.

Test suite

How it works

s.e%QssMc`Q]k`   Program. Input: Q
s.e%QssMc`Q]k`Q  Implicit input fill
 .e          `Q  Map over str(Q) with k as 0-indexed index:
        c`Q]k     Split str(Q) into two parts at index k
      sM          Convert both elements to integers
     s            Sum
   %Q             Q % that
s                Sum
                 Implicitly print

Haskell, 62 bytes

f x|m<-mod x=sum[m$div x(10^k)+m(10^k)|(k,_)<-zip[0..]$show x]

Defines a function f. See it pass all test cases.


Perl 6, 33 bytes

{sum $_ X%m:ex/^(.+)(.+)$/».sum}


{                  # bare block lambda with implicit parameter 「$_」


    $_             # the input

    X%             # cross modulus

    m :exhaustive /  # all possible ways to segment the input
    /».sum         # sum the pairs

Mathematica, 75 bytes


This uses pattern matching on the list of digits to extract all partitions of them into two parts. Each such partition into a and b is then replaced with


The notable thing here is that sums of lists of unequal length remain unevaluated, so e.g. if a is 1,2 and b is 3,4,5 then we first replace this with {1,2} + {3,4,5} + {}. The last term is there to ensure that it still remains unevaluated when we evenly split an even number of digits. Now the Map operation in Mathematica is sufficiently generalised that it works with any kind of expression, not just lists. So if we map FromDigits over this sum, it will turn each of those lists back into a number. At that point, the expression is a sum of integers, which now gets evaluated. This saves a byte over the more conventional solution Tr[FromDigits/@{{a},{b}}] which converts the two lists first and then sums up the result.


Actually, 16 15 bytes

Golfing suggestions welcome! Try it online!

Edit: -1 byte thanks to Teal pelican.



         Implicit input n.
;╗       Save a copy of n to register 0.
$l       Yield the number of digits the number has, len_digits.
r        Yield the range from 0 to len_digits - 1, inclusive.
`...`M   Map the following function over that range, with variable x.
  ╤        Yield 10**x.
  ╜        Push a copy of n from register 0.
  d        Push divmod(n, 10**x).
  +        Add the div to the mod.
  ╜        Push a copy of n from register 0.
  %        Vectorized modulo n % x, where x is a member of parition_sums.
         This function will yield a list of modulos.
Σ        Sum the results together.
         Implicit return.
  • \$\begingroup\$ If you move ╜% inside the functions section you don't need to use the ♀ and it'll save you 1 byte :D (;╗$lr╤╜d+╜%MΣ) \$\endgroup\$ Dec 12, 2016 at 15:46
  • \$\begingroup\$ @Tealpelican Thanks for the tip :D Let me know if you come up with any other golfing suggestions \$\endgroup\$
    – Sherlock9
    Dec 12, 2016 at 16:06

Prolog (SWI), 63 bytes

N+C+M:-N>C,N+C*10+T,M is N mod(N//C+N mod C)+T;M=0.

Try it online!

-1 thanks to Jo King

More interesting:

Prolog (SWI), 104 bytes

N-M:-findall(N mod(A+B),(N/C,append(X,Y,C),X\=[],Y\=[],A/X,B/Y),L),sumlist(L,M).

Try it online!

  • \$\begingroup\$ N+10+M can be N+1+M. I also tried to combine the functions, but it ended up the same size \$\endgroup\$
    – Jo King
    Sep 2, 2022 at 8:01

Ruby, 64 bytes


Takes the input as a string

  • \$\begingroup\$ Unfortunately, Ruby interprets integer literals beginning with 0 as octal, meaning this fails for the last test case. Here's a 78-byte solution addressing that. \$\endgroup\$
    – benj2240
    Feb 16, 2018 at 3:14

Befunge, 101 96 bytes

\<$>>\1-:v ^<^!:-1 

Try it online!


&              Read n from stdin.
100p           Initialise the current digit number to 1.

               -- The main loop starts here --

:::            Make several duplicates of n for later manipulation.

v+55g00<       Starting top right, and ending bottom right, this
>>\1-:v          routine calculates 10 to the power of the
^*+55\_$         current digit number.

%              The modulo of this number gives us the initial digits.
\              Swap another copy of n to the top of the stack.

_\55+*v        Starting bottom left and ending top left, this
^!:-1\<          is another calculation of 10 to the power of
00g55+^          the current digit number.

/              Dividing by this number gives us the remaining digits.
+              Add the two sets of digits together.
%              Calculate n modulo this sum.
\              Swap the result down the stack bringing n back to the top.

00g1+          Add 1 to the current digit number.
:9`#v_         Test if that is greater than 9.
00p            If not, save it and repeat the main loop.

               -- The main loop ends here --

$$             Clear the digit number and N from the stack.
++++++++       Sum all the values that were calculated.
.@             Output the result and exit.

APL, 29 bytes


⎕IO must be 1. Explanation (I'm not good at explaining, any imporvements to this are very welcome):

{                           } - Function (Monadic - 1 argument)
                           ⍵  - The argument to the function
                          ⍕   - As a string
                        R←    - Stored in R
                      1↓      - All except the first element
                    ⍳⍴        - 1 to the length
      {           }           - Another function
               ⍵↓R            - R without ⍵ (argument of inner function) leading digits
              ⍎               - As a number
             +                - Plus
       (    )                 - Grouping
         ⍵↑R                  - The first ⍵ digits of R
        ⍎                     - As a number
                   ¨          - Applied to each argument
   ⍵|⍨                        - That modulo ⍵ (outer function argument)
 +/                           - Sum
  • \$\begingroup\$ There, it's done. \$\endgroup\$
    – Adalynn
    Dec 12, 2016 at 21:35

C#, 67 bytes

n=>{long d=n,s=0,p=1;while(d>9)s+=n%((d/=10)+n%(p*=10));return s;};

Full program with ungolfed, explained method and test cases:

using System;

public class Program
    public static void Main()
        Func<long,long> f=
            long d = n, // stores the initial number
                 r,         // remainder, stores the last digits
                 s = 0, // the sum which will be returned
                 p = 1; // used to extract the last digit(s) of the number through a modulo operation
            while ( d > 9 ) // while the number has more than 1 digit
                d /= 10;    // divides the current number by 10 to remove its last digit
                p *= 10;    // multiplies this value by 10
                r = n % p;  // calculates the remainder, thus including the just removed digit
                s += n % (d + r);   // adds the curent modulo to the sum
            return s;   // we return the sum
        // test cases:
        Console.WriteLine(f(47852)); //5842
        Console.WriteLine(f(13));   // 1
        Console.WriteLine(f(111));  // 6
        Console.WriteLine(f(12345));    // 2097
        Console.WriteLine(f(54321));    // 8331
        Console.WriteLine(f(3729105472));   // 505598476

Attache, 48 bytes


Try it online!


                                          Digits    convert input to digits
    {                                   }@          call lambda with digits as argument
                      SplitAt[_,1..#_-1]            split digits at each partition
              Map[N]=>                              Map N to two-deep elements
          Sum@                                      Takes the sum of each sub array
     N[_]%                                          convert digits to int and vector mod
Sum@                                                sum the resultant array

Pip, 15 bytes


Try It Online!


              a  Program argument: the input number
            ,#   Range(length)
          MS     Map this function to each of those numbers and sum the results:
     a^@_          Split the input number at that index
  $+(    )         Sum the two halves
a%                 Take the input number modulo the sum

Using range(len) as the list of indices to split on means that in addition to 4785+2, 478+52, 47+852, and 4+7852, we also have +47852. However, a number modulo itself is 0, which doesn't change the sum, so that's no problem.


Clojure, 91 81 bytes

Edit: this is shorter as it declares an anonymous function (fn[v](->> ...)) and not using ->> macro although it was easier to read and call that way.

(fn[v](apply +(map #(mod v(+(quot v %)(mod v %)))(take 10(iterate #(* 10 %)1)))))


(defn s[v](->> 1(iterate #(* 10 %))(take 10)(map #(mod v(+(quot v %)(mod v %))))(apply +)))

Generates a sequence of 1, 10, 100, ... and takes first 10 items (assuming input values are less than 10^11), maps to modulos as specified in specs and calculates the sum. Long function names makes this solution quite long but at least even the golfed version should be quite easy to follow.

First I tried juggling strings around but it required tons of boilerplate.


Racket 134 bytes

(let((s(number->string n))(h string->number)(g substring))(for/sum((i(range 1(string-length s))))(modulo n(+(h(g s 0 i))(h(g s i))))))


(define (f n)
  (let ((s (number->string n))
        (h string->number)
        (g substring))
    (for/sum ((i (range 1 (string-length s))))
      (modulo n (+ (h (g s 0 i)) (h (g s i)))))))


(f 47852)
(f 13)
(f 111)
(f 12345)
(f 54321)
(f 3729105472)


  • \$\begingroup\$ So many close parens ... :D \$\endgroup\$ Dec 12, 2016 at 15:57
  • \$\begingroup\$ It's not as difficult as it appears. \$\endgroup\$
    – rnso
    Dec 12, 2016 at 16:34

Ruby 45 Bytes


This is a really neat solution. It is technically correct, but it is super inefficient. It would be much more efficient to write q.to_s.size.times{...}. We use q.times because it saves characters, and the extra number of times it goes through the proc the expression just evaluates to zero.

  • \$\begingroup\$ Sorry! This is a 45 byte solution written in ruby. I've edited the post to reflect that. \$\endgroup\$ Dec 20, 2016 at 20:03
  • \$\begingroup\$ 46 byte runner-up: ->q{(0..q).reduce{|s,x|p=10**x;s+q%(q/p+q%p)}} \$\endgroup\$ Dec 20, 2016 at 20:25

R, 50 bytes


Try it online!


SNOBOL4 (CSNOBOL4), 92 bytes

S	N LEN(X) . L REM . R	:F(O)
	S =S + REMDR(N,L + R)
	X =X + 1	:(S)

Try it online!


K (oK), 31 bytes



Try it online!




Most of the code goes to building up the lists out of the input. Am wondering if there is a trick I'm missing.

(4 7852
47 852
478 52
4785 2)

Full breakdown:

+/(+/+.:''1_(0,'!#$x)_\:$x)!'x: / the solution
                             x: / store input as variable x
                           !'   / mod (!) each-both (')
  (                       )     / do this together
                        $x      / string x
                     _\:        / cut (_) each-left (\:)
            (       )           / do this together
                  $x            / string x
                 #              / count length of the string
                !               / til, range 0..n-1
             0,'                / prepend 0 to each
          1_                    / drop the first one
      .:''                      / value (.:) each-each (nested)
     +                          / flip rows and columns
   +/                           / sum up the columns
+/                              / sum up results of the modulo

Vyxal s, 14 11 bytes

-3 bytes thanks to Lyxal


Try it online, or verify all test cases.


ż            # Range from 1 to the length of the input
  v          # Map each of those to
 ð           # A space
   Ṁ         # Inserted into the input at the nth index (resulting in ["4 7852", "47 852", "478 52", ...])
    Ṫ        # Remove the last one (which has doesn't space)
     ƛ       # On each string with the inserted space:
      Ė      #   Evaluate it as Vyxal code (so push both numbers in the string to the stack)
       +     #   Add them
        ?    #   Push the input
         $   #   Swap the top two items
          %  #   Modulo
             # Sum with the s flag

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