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!

What's the maximum input number to expect? With 32-bit integers in R this challenge can't be solved accurately past n>21 — Billywob, Nov 23, 2016 at 15:22
@Billywob For R then you will only need to go to 20. I shall edit question to reflect this — george, Nov 23, 2016 at 15:23

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

C, 63 60 bytes

-3 byte for do...while loop.

i;f(n){i=n;while(--n)i*=n;do n+=i%10;while(i/=10);return n;}

Ungolfed and usage:

i;
f(n){
 i=n;
 while(--n)
  i*=n;
 do
  n+=i%10;
 while(i/=10);
 return n;
}

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

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 23, 2016 at 15:20

Karl Napf

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\$\begingroup\$ Do we define f (n) as int by default? \$\endgroup\$
– Mukul Kumar
Nov 25, 2016 at 14:09
\$\begingroup\$ @MukulKumar this is standard in C, if there is no type then int is assumed. \$\endgroup\$
– Karl Napf
Nov 25, 2016 at 17:14

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Adam J · Accepted Answer · 2016-11-23 15:27:43Z

1

Q/KDB+ 41 Bytes

sum{:"I"$string x}over'string prd 1+til n

Breakdown:

string prd 1+til n

Get the product from 1 to n

{:"I"$string x}

Function that accepts a string x and returns it converted to an integer.

over'

Iterate over the argument to it's right to the function on the left, passing each item in individually.

sum

Sum up the numbers output by the function at the end.

answered Nov 23, 2016 at 15:27

Adam J

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user59178 · Accepted Answer · 2016-11-23 15:47:52Z

1

PHP, 44 bytes

<?=array_sum(str_split(gmp_fact($argv[1])));

Well it's not clever, but it works. That said it turned out better than I thought (I thought I'd need to use gmp_strval() too).

answered Nov 23, 2016 at 15:47

user59178

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Pyke, 3 bytes (old version)

SBs

Explanation:

SB  -  product(range(1, input+1))
  s - digit_sum(^)

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 23, 2016 at 17:10

Blue

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Groovy, 61 bytes

{"${(1..it).inject{i,r->i*r}}".collect{0.parseInt(it)}.sum()}

Groovy doesn't even have factorial built-ins.

"${(1..it).inject{i,r->i*r}}" - Compute factorial as string.

.collect{0.parseInt(it)} - Turn the string into an array of integers.

.sum() - Sum 'em.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 23, 2016 at 19:45

Magic Octopus Urn

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Mama Fun Roll · Accepted Answer · 2016-11-24 00:48:41Z

1

Wonder, 22 bytes

@sum <>""prod rng1+1#0

This only works for inputs <= 21.

Usage:

(@sum <>""prod rng1+1#0)10

Explanation

Increment argument, tail-exclusive range from 1 to result, product, split over empty string, and sum.

answered Nov 24, 2016 at 0:48

Mama Fun Roll

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CodeMonkey · Accepted Answer · 2016-11-24 13:32:34Z

1

F#, 80 bytes (74bytes for 32bit integer)

let rec s=function|0L->0L|y->y%10L+s(y/10L)
let g x={1L..x}|>Seq.fold (*) 1L|>s

F#, 98 bytes (96bytes for 32bit integer)

let f x=([1L..x]|>List.fold (*) 1L).ToString().ToCharArray()|>Array.fold (fun a c-> a+int(c)-48) 0

Test with

f 10L
g 10L

Thanks to ais523 for the anonymous match function and formatting help.

edited Nov 24, 2016 at 13:32

answered Nov 24, 2016 at 12:34

CodeMonkey

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2

\$\begingroup\$ Hi, and welcome to the site! Your answers are actually slightly better than you thought; unless it's vital to your program, you can remove the newline at the end of a file and save a byte that way. (You can produce the nicely formatted headings we use elsewhere by starting the heading with a # sign.) Also, I think F# has the function notation for creating an anonymous match block (e.g. you can use let rec s=function|0L rather than let rec s x=match x with|0L); that should help you get your code a bit shorter. \$\endgroup\$
– user62131
Nov 24, 2016 at 13:22
\$\begingroup\$ Thank you, it's the first time I saw the anonymous match. I select the functions in my code editor which told me 81 characters selected without the newline. I'm wondering if I got it wrong. \$\endgroup\$
– CodeMonkey
Nov 24, 2016 at 13:35

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

C#, 119 116 Bytes

Edit: Saved 3 Bytes thanks to @TheLethalCoder

Golfed:

long F(int n){long f=1;int s=0;for(;n>1;)f*=n--;foreach(var c in f.ToString().ToList())s+=int.Parse(c+"");return s;}

Ungolfed:

public long F(int n)
{
  long f = 1;
  int s = 0;
  for (; n > 1;)
    f *= n--;
  foreach (var c in f.ToString().ToList())
    s += int.Parse(c + "");
  return s;
}

Tried both a long and an Int64 and it won't take higher than n=20, ultimately went with long because it is shorter...

Testing:

  Console.WriteLine(new DigitalSumFactorial().F(10));
  //27
  Console.WriteLine(new DigitalSumFactorial().F(19));
  //45
  Console.WriteLine(new DigitalSumFactorial().F(20));
  //54

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 23, 2016 at 23:33

Pete Arden

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\$\begingroup\$ You can compile to a Func<int, long> to save bytes, the for loop can be simplified to for (; i > 1;) and then set i = n outside the loop above and post decrement i in the loop i.e. f *= i--;. I think the ToList() isn't needed although I haven't tried it and you will need to add using System.Linq; into the byte count for the ForEach part \$\endgroup\$
– TheLethalCoder
Nov 24, 2016 at 15:53
\$\begingroup\$ int.Parse(c+"") can be replaced with c-30 I think... \$\endgroup\$
– TheLethalCoder
Nov 24, 2016 at 16:18
\$\begingroup\$ Its 48 not 30 I read the hex code by accident, also you can't remove the ToList, putting all of that together is 113 bytes including the 18 for the using statement: using System.Linq;n=>{long f=1;int s=0,i=n;for(;i>1;)f*=i--;f.ToString().ToList().ForEach(c=>s+=c-48);return s;}; \$\endgroup\$
– TheLethalCoder
Nov 24, 2016 at 16:27
\$\begingroup\$ @TheLethalCoder Thanks for the tips :) I've had another play around with it and saved 3 Bytes :) \$\endgroup\$
– Pete Arden
Nov 24, 2016 at 18:11

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Lithp, 73 bytes

#N::((sum (map (split (+ "" (prod (seq 1 N))) "") #C::((- (asc C) 48)))))

I'm not sure if I should be counting import calls. Here is how you would use the above snippet:

(
    (import "lists")
    (def f #N::((sum (map (split (+ "" (prod (seq 1 N))) "") #C::((- (asc C) 48))))))
    (print (f 10))
)

It works by calculating the factorial using prod/1, converts it to a string by doing (+ "" value), splits and maps the string to get the numeric value of each number (ASCII code - 48.) Lastly, we sum/1 the resulting list.

RaviRavioli · Accepted Answer · 2018-02-22 15:10:32Z

1

Pyt, 3 2 bytes

!Ś

Explanation:

        Implicit input
!       Get factorial
 Ś      Sum the digits

Try it online!

edited Feb 22, 2018 at 15:10

RaviRavioli

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answered Dec 27, 2017 at 20:42

mudkip201

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Japt, 3 bytes

Êìx

Try it

Get the factorial, split to a digit array and reduce by addition.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Feb 22, 2018 at 15:17

Shaggy

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sergiol · Accepted Answer · 2018-04-24 16:46:36Z

1

Tcl, 78 bytes

proc S n {proc F n {expr $n?($n)*\[F $n-1]:1}
expr [join [split [F $n] ""] +]}

Try it online!

edited Apr 24, 2018 at 16:46

answered Apr 23, 2018 at 15:23

sergiol

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mkst · Accepted Answer · 2018-07-08 21:33:56Z

1

K (ngn/k), 10 bytes

Solution:

+/10\*/1+!

Try it online!

Explanation to follow.

answered Jul 8, 2018 at 21:33

mkst

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Jakob · Accepted Answer · 2018-07-27 19:43:18Z

Java 10, 70 bytes

A lambda from long to int. Breaks for input over 20.

n->{var f=n;while(n>1)f*=--n;return(f+"").chars().map(c->c-48).sum();}

Try It Online

Java 10, 163 bytes

Fully arbitrary precision. A lambda from BigInteger to BigInteger.

n->{var i=n;var f=n;while(n.compareTo(n.ONE)>0)f=f.multiply(n=n.subtract(n.ONE));return(f+"").chars().mapToObj(c->i.valueOf(c-48)).reduce(n.ZERO,(a,b)->a.add(b));}

Try It Online

Ungolfed

n -> {
    var i = n;
    var f = n;
    while (n.compareTo(n.ONE) > 0)
        f = f.multiply(n = n.subtract(n.ONE));
    return (f + "").chars()
        .mapToObj(c -> i.valueOf(c - 48))
        .reduce(n.ZERO, (a, b) -> a.add(b))
    ;
}

Xcali · Accepted Answer · 2020-12-22 16:52:42Z

1

Perl 5 `-MList::Util=sum -MMath::BigInt -p`, 43 bytes

Supports arbitrarily large factorials.

$_=sum(Math::BigInt->new($_)->bfac()=~/./g)

Try it online!

Perl 5 `-MList::Util=sum,product -p`, 27 bytes

Cannot handle extremely large factorials due to overflow.

$_=product 1..$_;$_=sum/./g

Try it online!

answered Dec 22, 2020 at 16:52

Xcali

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user · Accepted Answer · 2020-12-22 18:25:52Z

1

Scala, 30 bytes

1.to(_).product+""map(_-48)sum

Try it online!

And a 34 byte solution without postfix syntax:

x=>(""+1.to(x).product:\0)(_+_-48)

Try it in Scastie

answered Dec 22, 2020 at 18:25

user

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south · Accepted Answer · 2022-11-22 06:24:27Z

1

Fig, \$3\log_{256}(96)\approx\$ 2.469 bytes

Sra

Try it online!

Sra
  a  # (num) range [1, a]
 r   # (lst) product
S    # (num) digit sum

answered Nov 22, 2022 at 6:24

south

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Nati · Accepted Answer · 2016-11-24 14:41:17Z

0

Java8 - 112 Chars

(First time here...)

String.valueOf(LongStream.rangeClosed(2,i).reduce(1,(a,b)->a*b)).chars().map(Character::getNumericValue).sum();

answered Nov 24, 2016 at 14:41

Nati

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gggg · Accepted Answer · 2018-01-04 18:38:03Z

0

Julia 0.6, 25 bytes

Julia functions are generic across input types, and generate optimized code for each combination of input types. This function works with machine sized integers or arbitrary precision integers depending on the type of the argument. eg

f(10) = 27
f(50) = -97 # overflowed
f(big(50)) = 216 # no overflow, but slower due to use of BigInt

x->sum(digits(prod(1:x)))

Try it online!

answered Jan 4, 2018 at 18:38

gggg

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Steadybox · Accepted Answer · 2018-01-14 13:24:45Z

0

Clojure, 65 bytes

(fn[i](apply +(map #(-(int %)48)(str(apply *(range 1(inc i)))))))

Straightforward, character \0 has integer value of 48.

Easier to follow steps:

(defn f [i] (->> i inc (range 1) (apply *) str (map #(- (int %) (int \0)) (apply +)))

edited Jan 14, 2018 at 13:24

Steadybox

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answered Dec 15, 2016 at 17:31

NikoNyrh

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alephalpha · Accepted Answer · 2018-04-24 17:10:32Z

0

Pari/GP, 16 bytes

n->sumdigits(n!)

Try it online!

edited Apr 24, 2018 at 17:10

answered Jan 13, 2018 at 15:08

alephalpha

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Jonathan Frech · Accepted Answer · 2018-07-07 13:02:20Z

0

Python 3, 66 bytes

import math;print(sum(map(int,str(math.factorial(int(input()))))))

It gets factorial of number, splits it into an array, then prints it!

edited Jul 7, 2018 at 13:02

Jonathan Frech

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answered Jul 7, 2018 at 11:48

aryaman

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\$\begingroup\$ Welcome to PPCG! \$\endgroup\$
– Jonathan Frech
Jul 7, 2018 at 13:02

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Sundar R · Accepted Answer · 2018-07-07 19:23:41Z

MATL, 9 bytes

:p[]&V!Us

Try it online!

MATL port of Stewie Griffin's Octave answer. Works up to n=22, same as the Octave and R answers.

:p - factorial.
[]&V - string representation of the number, fully expanded (without the []&, larger numbers use the scientific notation which won't do for our purposes).
!U transpose and convert back to numbers, getting individual digits.
s - sum those digits.

The []& part can be removed for -3 bytes, but then the range is further limited to only work up to 17, instead of 22.

I could stretch the input range to 23 with a couple of tricks, but that costs an additional 20 bytes:

:1w5X2Y%"@*t10\~?10/]][]&V!Us

(basically, use the uint64 data type instead of the usual double, multiply by each number upto the given number, but divide away 10 whenever our product is divisible by it (since trailing zeros add nothing to the digit sum).)

That seems to be as far as we can go with MATLAB/MATL without doing an ad-hoc implementation of bignum in the program.

lyxal · Accepted Answer · 2021-03-16 11:58:21Z

0

Vyxal, 2 bytes

‼∑

Try it Online!

A very literal interpretation of the challenge: sum_of_digits_of(factorial(input))

answered Mar 16, 2021 at 11:58

lyxal

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cillianreilly · Accepted Answer · 2022-11-18 01:38:26Z

0

q, 19 bytes

sum 10 vs prd 1+til

k, 11 bytes

+/10\:*/1+!

prd 1+til generate the factorial of input

10 vs convert factorial to digits

answered Nov 18, 2022 at 1:38

cillianreilly

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Jordan · Accepted Answer · 2022-11-21 22:28:42Z

0

J-uby, 21 bytes

:+|:/&:*|:digits|:sum

Attempt This Online!

Explanation

:+ | :/ & :* | :digits | :sum

:+ |                           # Range 1..n, then
     :/ & :* |                 # Reduce with product, then
               :digits | :sum  # Sum digits

edited Nov 21, 2022 at 22:28

answered Nov 17, 2022 at 22:48

Jordan

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Yousername · Accepted Answer · 2022-11-23 23:40:01Z

0

TI-Basic, 24 bytes (non-competing)

sum(10fPart(.1seq(int(Ans!/₁₀^(I)),I,0,int(log(Ans!

Only works for nonnegative integers up to 18, inclusive, and works up to 20 if you round to the nearest multiple of 9 when the input is greater than 5.

edited Nov 23, 2022 at 23:40

answered Nov 22, 2022 at 3:04

Yousername

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Factorial digit sum

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

57 Answers 57

C, 63 60 bytes

PHP, 44 bytes

Pyke, 3 bytes (old version)

Groovy, 61 bytes

Wonder, 22 bytes

Explanation

F#, 80 bytes (74bytes for 32bit integer)

F#, 98 bytes (96bytes for 32bit integer)

C#, 119 116 Bytes

Lithp, 73 bytes

Pyt, 3 2 bytes

Japt, 3 bytes

Tcl, 78 bytes

K (ngn/k), 10 bytes

Java 10, 70 bytes

Java 10, 163 bytes

Ungolfed

Perl 5 -MList::Util=sum -MMath::BigInt -p, 43 bytes

Perl 5 -MList::Util=sum,product -p, 27 bytes

Scala, 30 bytes

Fig, \$3\log_{256}(96)\approx\$ 2.469 bytes

Java8 - 112 Chars

Julia 0.6, 25 bytes

Clojure, 65 bytes

Pari/GP, 16 bytes

Python 3, 66 bytes

MATL, 9 bytes

Vyxal, 2 bytes

q, 19 bytes

k, 11 bytes

J-uby, 21 bytes

Explanation

TI-Basic, 24 bytes (non-competing)

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