Factorial digit sum

Question

The challenge is to calculate the digit sum of the factorial of a number.

---

Example

Input: 10
Output: 27

10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27

You can expect the input to be an integer above 0. Output can be of any type, but the answer should be in the standard base of the coding language.

---

Test cases:

10    27
19    45
469   4140
985   10053

N.B. Some languages can not support large numbers above 32-bit integers; for those languages you will not be expected to calculate large factorials.

OEIS link here thanks to Martin Ender

---

This is code-golf, so shortest code in characters wins!

Answer 1

05AB1E, 3 bytes

!SO

Try it online!

!   Factorial.
 S  Push characters separately.
  O Sum.

Answer 2

Jelly, 3 bytes

!DS

Try it online!

Does what you expect:

!    Factorial.
 D   Decimal digits.
  S  Sum.

Answer 3

Mathematica, 21 bytes

Tr@IntegerDigits[#!]&

Answer 4

C++11, 58 bytes

As unnamed lambda modifying its input:

[](int&n){int i=n;while(--n)i*=n;do n+=i%10;while(i/=10);}

One of the rare cases when my C++ code is shorter than the C code.

If you want to suppport larger cases, switch to C++14 and use:

[](auto&n){auto i=n;while(--n)i*=n;do n+=i%10;while(i/=10);}

and supply the calling argument with ull suffix.

Usage:

auto f=
[](int&n){int i=n;while(--n)i*=n;do n+=i%10;while(i/=10);}
;

main() {
  int n=10;
  f(n);
  printf("%d\n",n);
}

Answer 5

Ruby, 63 61 53 38 bytes

New approach thanks to manatwork:

->n{eval"#{(1..n).reduce:*}".chars*?+}

Old:

->n{(1..n).reduce(:*).to_s.chars.map(&:hex).reduce:+}

• -3 bytes thanks to Martin Ender
• -5 bytes thanks to G B

Answer 6

Pyth, 7 6 bytes

Thanks to @Kade for saving me a byte

sj.!QT

Try it online!

This is my first time using Pyth, so I'm sure that my answer could be golfed quite a bit.

Explanation:

s Sum
  j the digits of
    .! the factorial of
      Q the input
    T in base 10

Answer 7

Haskell, 41 40 bytes

f x=sum$read.pure<$>(show$product[1..x])

Usage example: f 985 -> 10053.

Make a list from 1to x, calculate the product of the list elements, turn it into its string representation, turn each character into a number and sum them.

Edit: @Angs saved a byte. Thanks!

Answer 8

Python, 54 bytes

f=lambda n,r=1:n and f(n-1,r*n)or sum(map(int,str(r)))

repl.it

Answer 9

R, 58 53 bytes

Edit: Saved one byte thanks to @Jonathan Carroll and a couple thanks to @Micky T

sum(as.double(el(strsplit(c(prod(1:scan()),""),""))))

Unfortunately, with 32-bit integers, this only works for n < 22. Takes input from stdin and outputs to stdout.

If one would like higher level precision, one would have to use some external library such as Rmpfr:

sum(as.numeric(el(strsplit(paste(factorial(Rmpfr::mpfr(scan()))),""))))

Answer 10

Pip, 8 bytes

$+$*++,a

Try it online!

Explanation

      ,a    # range
    ++      # increment
  $*        # fold multiplication
$+          # fold sum

Answer 11

CJam, 8 bytes

rim!Ab:+

Try it online!

Explanation

r   e# Read input.
i   e# Convert to integer.
m!  e# Take factorial.
Ab  e# Get decimal digits.
:+  e# Sum.

Answer 12

APL (Dyalog Extended), 8 bytes

-3 thanks to @Razetime - i forgot to drop a paren lol

+/⍎¨∘⍕∘!

Explanation:

+/⍎¨∘⍕∘!
       !  ⍝ factorial of ⍵
     ⍕    ⍝ convert to string
  ⍎¨      ⍝ produce digit vector
+/        ⍝ sum all digits

Has some problems with factorial of 900, this code will work fine for ⍵>40:

b←({⍵⊣⍵.⎕CY'dfns'}⎕NS⍬).big
↑(+b/(⍎¨∘⍕∘(↑(×b/⍳))))

Try it online!

Answer 13

Brachylog, 5 bytes

$!@e+

Try it online!

Explanation

Basically the described algorithm:

$!       Take the factorial of the Input
  @e     Take the elements of this factorial (i.e. its digits)
    +    Output is the sum of those elements

Answer 14

Java 7, 148 bytes

int s=1,ret=0;while(i>1){s=s*i; i--;}String y=String.valueOf(s);for(int j=0;j<y.length();j++){ret+=Integer.parseInt(y.substring(j,j+1));}return ret;

Answer 15

Ruby, 63 60 53 51 bytes

->n{a=0;f=(1..n).reduce:*;f.times{a+=f%10;f/=10};a}

Thanks to Martin for golfing help.

Answer 16

Pushy, 4 bytes

fsS#

Give input on the command line: $ pushy facsum.pshy 5. Here's the breakdown:

f      % Factorial of input
 s     % Split into digits
  S    % Push sum of stack
   #   % Output

Answer 17

Octave, 30 bytes

@(n)sum(num2str(prod(1:n))-48)

Calculates the factorial by taking the product of the list [1 2 ... n]. Converts it to a string and subtracts 48 from all elements (ASCII code for 0). Finally it sums it up :)

Answer 18

bash (seq,bc,fold,jq), 34 33 bytes

Surely not the most elegant but for the challenge

seq -s\* $1|bc|fold -1|jq -s add

Answer 19

C, 58 bytes

This is not perfect. Only works ones because a have to be -1 in start. The idea is to use two recursive function in one function. It was not as easy as I first thought.

a=-1;k(i){a=a<0?i-1:a;return a?k(i*a--):i?i%10+k(i/10):0;}

Usage and understandable format:

a = -1;
k(i){
   a = a<0 ? i-1 : a;
   return a ? k(i*a--) : i? i%10+k(i/10) :0;
}

main() {
   printf("%d\n",k(10));
}

Edit: I found metode that let use this function multiple time but then length is 62 bytes.

a,b;k(i){a=b?a:i+(--b);return a?k(i*a--):i?i%10+k(i/10):++b;}

Answer 20

Perl 6, 21 bytes

{[+] [*](2..$_).comb}

Expanded:

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

  [+]           # reduce the following with ｢&infix:<+>｣

    [*](        # reduce with ｢&infix:<*>｣
      2 .. $_   # a Range that include the numbers from 2 to the input (inclusive)
    ).comb      # split the product into digits
}

Answer 21

Cubix, 33 32 bytes

u*.$s.!(.01I^<W%NW!;<,;;q+p@Opus

Net form:

      u * .
      $ s .
      ! ( .
0 1 I ^ < W % N W ! ; <
, ; ; q + p @ O p u s .
. . . . . . . . . . . .
      . . .
      . . .
      . . .

Try it online!

Notes

• Works with inputs up to and including 170, higher inputs result in an infinite loop, because their factorial yields the Infinity number (technically speaking, its a non-writable, non-enumerable and non-configurable property of the window object).
• Accuracy is lost for inputs 19 and up, because numbers higher than 2⁵³ (= 9 007 199 254 740 992) cannot be accurately stored in JavaScript.

Explanation

This program consists of two loops. The first calculates the factorial of the input, the other splits the result into its digits and adds those together. Then the sum is printed, and the program finishes.

Start

First, we need to prepare the stack. For that part, we use the first three instructions. The IP starts on the fourth line, pointing east. The stack is empty.

      . . .
      . . .
      . . .
0 1 I . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
      . . .
      . . .
      . . .

We will keep the sum at the very bottom of the stack, so we need to start with 0 being the sum by storing that on the bottom of the stack. Then we need to push a 1, because the input will initially be multiplied by the number before it. If this were zero, the factorial would always yield zero as well. Lastly we read the input as an integer.

Now, the stack is [0, 1, input] and the IP is at the fourth line, the fourth column, pointing east.

Factorial loop

This is a simple loop that multiplies the top two elements of the stack (the result of the previous loop and the input - n, and then decrements the input. It breaks when the input reaches 0. The $ instruction causes the IP to skip the u-turn. The loop is the following part of the cube. The IP starts on the fourth line, fourth column.

      u * .
      $ s .
      ! ( .
. . . ^ < . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
      . . .
      . . .
      . . .

Because of the ^ character, the IP starts moving north immediately. Then the u turns the IP around and moves it one to the right. At the bottom, there's another arrow: < points the IP back into the ^. The stack starts as [previousresult, input-n], where n is the number of iterations. The following characters are executed in the loop:

*s(
*   # Multiply the top two items
    #   Stack: [previousresult, input-n, newresult]
 s  # Swap the top two items
    #   Stack: [previousresult, newresult, input-n]
  ( # Decrement the top item
    #   Stack: [previousresult, newresult, input-n-1]

Then the top of the stack (decreased input) is checked against 0 by the ! instruction, and if it is 0, the u character is skipped.

Sum the digits

The IP wraps around the cube, ending up on the very last character on the fourth line, initially pointing west. The following loop consists of pretty much all remaining characters:

      . . .
      . . .
      . . .
. . . . . W % N W ! ; <
, ; ; q + p @ O p u s .
. . . . . . . . . . . .
      . . .
      . . .
      . . .

The loop first deletes the top item from the stack (which is either 10 or 0), and then checks what is left of the result of the factorial. If that has been decreased to 0, the bottom of the stack (the sum) is printed and the program stops. Otherwise, the following instructions get executed (stack starts as [oldsum, ..., factorial]):

N%p+q;;,s;
N          # Push 10
           #   Stack: [oldsum, ..., factorial, 10]
 %         # Push factorial % 10
           #   Stack: [oldsum, ..., factorial, 10, factorial % 10]
  p        # Take the sum to the top
           #   Stack: [..., factorial, 10, factorial % 10, oldsum]
   +       # Add top items together
           #   Stack: [..., factorial, 10, factorial % 10, oldsum, newsum]
    q      # Send that to the bottom
           #   Stack: [newsum, ..., factorial, 10, factorial % 10, oldsum]
     ;;    # Delete top two items
           #   Stack: [newsum, ..., factorial, 10]
       ,   # Integer divide top two items
           #   Stack: [newsum, ..., factorial, 10, factorial/10]
        s; # Delete the second item
           #   Stack: [newsum, ..., factorial, factorial/10]

And the loop starts again, until factorial/10 equals 0.

Answer 22

C, 47 bytes

f(n,a){return n?f(n-1,a*n):a?a%10+f(0,a/10):0;}

usage:

f(n,a){return n?f(n-1,a*n):a?a%10+f(0,a/10):0;}
main() {
  printf("answer: %d\n",f(10,1));
}

Answer 23

Brachylog (v2), 3 bytes

ḟẹ+

Try it online!

Same "algorithm" as the v1 answer by @Fatalize, just with better encoding.

Answer 24

Python, 57 bytes

import math
lambda n:sum(map(int,str(math.factorial(n))))

Try it online

Answer 25

Batch, 112 bytes

@set/af=1,t=0
@for /l %%i in (1,1,%1)do @set/af*=%%i
:g
@set/at+=f%%10,f/=10
@if %f% gtr 0 goto g
@echo %t%

Conveniently set/a works on a variable's current value, so it works normally inside a loop. Only works up to 12 due to the limitations of Batch's integer type, so in theory I could save a byte by assuming f<1e9:

@set/af=1,t=0
@for /l %%i in (1,1,%1)do @set/af*=%%i
@for /l %%i in (1,1,9)do @set/at+=f%%10,f/=10
@echo %t%

But that way lies madness... I might as well hard-code the list in that case (97 bytes):

@call:l %1 1 1 2 6 6 3 9 9 9 27 27 36 27
@exit/b
:l
@for /l %%i in (1,1,%1)do @shift
@echo %2

Answer 26

JavaScript (ES6), 50 bytes

f=(n,m=1,t=0)=>n?f(n-1,n*m):m?f(n,m/10|0,t+m%10):t

Only works up to n=22 due to floating-point accuracy limitations.

Answer 27

Befunge 93, 56 54 bytes

Saved 2 bytes do to using get instead of quotes. This let me shift the top 2 lines over 1, reducing unnecessary white space.

Try it online!

&#:<_v#:-1
: \*$<:_^#
g::v>91+%+00
_v#<^p00</+19
@>$$.

Explanation:

&#:<                Gets an integer input (n), and reverses flow direction
&#:< _v#:-1         Pushes n through 0 onto the stack (descending order)

:  \*$<:_^#         Throws the 0 away and multiplies all the remaining numbers together

(reorganized to better show program flow):
vp00< /+19 _v#<    Stores the factorial at cell (0, 0). Pushes 3 of whatever's in
> 91+%+ 00g ::^    cell (0, 0). Pops a, and stores a / 10 at (0, 0),
                   and adds a % 10 to the sum.

@>$$.              Simply discards 2 unneeded 0s and prints the sum.

Answer 28

Javascript ES6 - 61 54 Bytes

n=>eval(`for(j of''+(a=_=>!_||_*a(~-_))(n,t=0))t-=-j`)

EDIT: Thank you Hedi and ETHproductions for shaving off 7 bytes. I'll have to remember that t-=-j trick.

Answer 29

AHK, 60 bytes

a=1
Loop,%1%
a*=A_Index
Loop,Parse,a
b+=A_LoopField
Send,%b%

AutoHotkey doesn't have a built-in factorial function and the loop functions have long names for their built-in variables. The first loop is the factorial and the second is adding the digits together.

Answer 30

J, 12 11 bytes

Saved 1 byte thanks to cole!

1#.10#.inv!

This simply applies sum (1#.) to the digits (using inverse inv of base conversion #. with a base of 10) of the factorial (!) of the argument.

Test cases

Note: the last two test cases are bigints, as marked by a trailing x.

   f=:10#.inv!
   (,. f"0) 10 19 469x 985x
 10    27
 19    45
469  4140
985 10053