The Non-Zero Digital Product Challenge

Question

^{Originally the Multiplicative digital root}

Challenge

^{^{Basically do what the title says}}

Method

Given a positive integer 1 <= N <= 100000000 through one of our standard input methods, multiply every digit together, ignoring zeroes.

Ex: Take a number, say 361218402:

3 * 6 = 18
18 * 1 = 18
18 * 2 = 36
36 * 1 = 36
36 * 8 = 288
288 * 4 = 1152
1152 * 1 ^{(ignore zeroes or turn them into ones)} = 1152
1152 * 2 = 2304

The output for 361218402 is 2304

Note: Treat the digit 0 in numbers as 1.

Test Cases

1 => 1
10 => 1
20 => 2
100 => 1
999 => 729
21333 => 54
17801 => 56
4969279 => 244944
100000000 => 1

Standard Loopholes are disallowed, and this is code-golf, so shortest byte count wins!

Congrats to Jo King who got the bounty with his 70 byte brain-flak answer!

I'd rather call this non-zero digital product. "root" suggests it reduces to a single digit, which isn't always true here. — Erik the Outgolfer, Jan 15, 2018 at 20:46
Can we take input as a string? Or (pushing it) an array of digits? — Shaggy, Jan 15, 2018 at 21:19
@EriktheOutgolfer Yes, however, if you repeat the process enough times, it does appear to always go to a single digit. — DJMcMayhem, Jan 15, 2018 at 21:21
You can take quoted input, but no, you can't take a pre-parsed list of digits if that's what you're asking — qqq, Jan 15, 2018 at 21:21
If we have to support to a max of 100000000000 I suggest the test case 99999999999 => 31381059609, since it doesn't fit in a default 32-bit integer. Perhaps better would be to lower the maximum output to a 32-bit maximum (2147483647). — Kevin Cruijssen, Jan 16, 2018 at 8:27

Cristian Lupascu · Accepted Answer · 2018-01-15 20:30:17Z

1

Haskell, 36 bytes

f n=product[read[d]|d<-show n,d>'0']

Same byte count:

f n=product[max 1$read[d]|d<-show n]

Try it online!

answered Jan 15, 2018 at 20:30

Cristian Lupascu

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\$\begingroup\$ Also 36 \$\endgroup\$
– H.PWiz
Jan 15, 2018 at 20:56

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DJMcMayhem · Accepted Answer · 2018-01-15 20:41:11Z

1

Jelly, 6, 5 bytes

ṢU×\Ṁ

Try it online!

edited Jan 15, 2018 at 20:41

answered Jan 15, 2018 at 20:21

DJMcMayhem

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\$\begingroup\$ 5 mins, not bad! \$\endgroup\$
– qqq
Jan 15, 2018 at 20:26
\$\begingroup\$ DTịDP would save a byte, but there are better ways to strip zeroes or replace them with something else. \$\endgroup\$
– Dennis
Jan 15, 2018 at 20:34
5

\$\begingroup\$ Multiplicative root with ṢUṀ \$\endgroup\$
– Uriel
Jan 15, 2018 at 20:44

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Emigna · Accepted Answer · 2018-01-15 20:46:05Z

1

><>, 19 bytes

1i:&0(?n&c4*-:0=+*!

Try it online!

answered Jan 15, 2018 at 20:46

Emigna

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caird coinheringaahin g · Accepted Answer · 2018-01-15 20:55:48Z

1

Add++, 12 bytes

L,EDBFEZB]B*

Try it online!

answered Jan 15, 2018 at 20:55

caird coinheringaahin g♦

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Οurous · Accepted Answer · 2018-01-15 21:06:52Z

1

Clean, 55 bytes

import StdEnv
@n=prod[toInt c-48\\c<-:toString n|c>'0']

Try it online!

answered Jan 15, 2018 at 21:06

Οurous

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Graham · Accepted Answer · 2018-01-15 21:10:26Z

1

APL+WIN, 9 bytes

×/(⍎¨⎕)~0

Prompts for screen input as a character string, convert to digits, drop zeros and multiply all.

edited Jan 15, 2018 at 21:10

answered Jan 15, 2018 at 21:08

Graham

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Chas Brown · Accepted Answer · 2018-01-15 21:15:30Z

1

Python 2, 46 44 bytes

f=lambda n,v=1:n and f(n/10,n%10*v or v)or v

Try it online!

2 bytes thx to Lynn.

edited Jan 15, 2018 at 21:15

answered Jan 15, 2018 at 20:42

Chas Brown

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\$\begingroup\$ f(n/10,n%10*v or v) saves 2 bytes. \$\endgroup\$
– Lynn
Jan 15, 2018 at 21:08

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reffu · Accepted Answer · 2018-01-15 21:26:31Z

1

Forth (gforth), 62 bytes

: f 1 swap begin ?dup while 10 /mod -rot 1 max * swap repeat ;

Try it online!

answered Jan 15, 2018 at 21:26

reffu

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Uriel · Accepted Answer · 2018-01-15 21:42:26Z

1

APL (Dyalog), 11 7 bytes

4 bytes saved thanks to @EriktheOutgolfer

×/0~⍨⍎¨

Try it online!

edited Jan 15, 2018 at 21:42

answered Jan 15, 2018 at 20:39

Uriel

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\$\begingroup\$ 7 bytes (although I doubt you can take the number as a string, but then still 9 bytes) \$\endgroup\$
– Erik the Outgolfer
Jan 15, 2018 at 21:13

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Jonathan S. · Accepted Answer · 2018-01-15 21:43:12Z

1

Lua, 45 bytes

a=1(...):gsub("[^0]",load("a=a*..."))print(a)

Try it online!

answered Jan 15, 2018 at 21:43

Jonathan S.

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

1

Pyth, 10 9 Bytes

.U*s!BsZs

Try it!

Explanation

.U*s!BsZs
.U        Q     Fold over the implicit input string
      sZ        Convert each char to int
   s!B          Form the list [d,not d] and sum it, so that 0 becomes 1
  *     sb      Multiply by the prev product (implicit). s needed because 1st value is char

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 15, 2018 at 21:58

KarlKastor

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

1

Japt, 5 bytes

ì f ×

Test it here

Explanation

In order: split to an array of digits, filter to remove zeroes and reduce by multiplication.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 15, 2018 at 21:18

Shaggy

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Brad Gilbert b2gills · Accepted Answer · 2018-01-16 00:16:37Z

1

Perl 6, 19 bytes

{[*] grep +*,.comb}

Test it

Expanded:

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

  [*]        # reduce using &infix:«*»

    grep     # find the values
      +*,    # that aren't "0"
      .comb  # split ｢$_｣ into digits (returns single character strings)
}

answered Jan 16, 2018 at 0:16

Brad Gilbert b2gills

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Neil · Accepted Answer · 2018-01-16 00:43:37Z

1

Batch, 78 bytes

@set/ap=%2+0,p+=!p,d=%1%%10,p*=d+!d,n=%1/10
@if %1 gtr 0 %0 %n% %p%
@echo %2

Previous 80-byte solution:

@set/an=%1,p=1
:l
@set/ad=n%%10,p*=d+!d,n/=10
@if %n% gtr 0 goto l
@echo %p%

(This version outputs 1 for an input of zero, rather than failing.)

answered Jan 16, 2018 at 0:43

Neil

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ბიმო · Accepted Answer · 2018-01-16 08:46:34Z

1

Husk, 4 bytes

ΠfId

Try it online!

ΠfId  -- implicit input N, for example 1607
   d  -- digigts: [1,6,0,7]
 fI   -- filter by identity function: [1,6,7]
Π     -- product: 42

answered Jan 16, 2018 at 8:46

ბიმო

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ბიმო · Accepted Answer · 2018-01-16 08:58:49Z

dc, 40 bytes

Sad to see dc so underrepresented here.

This is essentially my first code golf post, so if I've misjudged how bytes should be counted, go easy on me! :)

?[1+]sc[10~rdZ1<a]dsax1[rd0=c*z1<b]dsbxp

You can run this in the terminal like so:

$ echo 361218402 | dc -e '?[1+]sc[10~rdZ1<a]dsax1[rd0=c*z1<b]dsbxp'
2304

Or you can try it online!

Explanation (probably too verbose):

? # Read and execute a line from standard input
  # (A number just goes on the stack)
[ # Begin a literal string (ended by a matched ']')
1 # Push 1 onto the stack (when this string is executed as a macro)
+ # Pop top two numbers, add them, push result
] # End literal string
s # Pop top of stack (the string) and store it in...
c # ...register c
[
10 # Push 10 (duh)
~ # Pop top two numbers, divide second popped by first popped, push quotient and then remainder
r # Reverse top two numbers of stack
d # Duplicate top of stack (pop and then push it twice)
Z # Pop top of stack and push its length in digits
1
< # Pop two numbers and compare; if first popped is less than second then...
a # ...run macro stored in register a (a recursive call, here)
] # End string
d # Duplicate top of stack
s # Store in...
a # ...register a
x # Execute the top of stack!
  # At this point the stack will consist of one digit entries, the digits of input
1
[
r # Reverse
d # Duplicate
0
= # If top two numbers are equal then...
c # ...run macro in register c
* # Pop two, multiply, push product
z # Push "height" of stack
1
< # If height of stack is greater than 1 then...
b # ...run macro b
]
d # Duplicate the string on the stack
sb # Store it as macro b
x # Execute!
p # Print top of stack

Always nice to see dc around :) I added a TIO link, hope you don't mind - feel free to rollback if you don't like my edit. — ბიმო, Jan 16, 2018 at 9:00
@BMO thanks! And I don’t know if this is maximally golfed; it’s pretty straightforward implementation. — Wildcard, Jan 16, 2018 at 9:01
You might be able to get rid of the first macro and 0=c with this tip, but I haven't tried myself. — ბიმო, Jan 16, 2018 at 9:10

Galen Ivanov · Accepted Answer · 2018-01-16 11:36:23Z

1

Red, 68 bytes

func[n][p: 1 while[n > 0][d: n // 10 n: n / 10 if d > 0[p: p * d]]p]

Try it online!

Since 100000000000 is beyond the integer range of the Red language, here's another approach that solves this issue, using string manipulation:

Red, 74 bytes

func[n][p: 1 foreach c to-string n[if c >#"0"[p: p * to-integer c - 48]]p]

Try it online!

answered Jan 16, 2018 at 11:36

Galen Ivanov

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david · Accepted Answer · 2018-01-16 12:08:36Z

1

Javascript, 58 bytes

f=a=>{b=1;for(i=0;a[i];i++){(a[i]>0)&&(b=+b*a[i])}alert(b)

Input via function(arg) outputs alert

edited Jan 16, 2018 at 12:08

answered Jan 16, 2018 at 11:34

david

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mkst · Accepted Answer · 2018-01-16 20:23:06Z

1

K (oK), 10 7 bytes

Solution:

*/1|.:'

Try it online!

Example:

*/1|.:'"4969279"
244944

Explanation:

Convert the input string to a list of integers, take max compared to 1 (inspiration from the J solution) and then multiply over the list.

Evaluation is performed right to left.

*/1|.:' / the solution
    .:' / value (.:) each (')
  1|    / or with 1 ( 0=>1, 1=>1, 2=>2 )
*/      / multiply (*) over (/)

edited Jan 16, 2018 at 20:23

answered Jan 15, 2018 at 21:23

mkst

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Giuseppe · Accepted Answer · 2018-01-16 21:12:47Z

1

SNOBOL4 (CSNOBOL4), 83 bytes

	DEFINE('O(N)')
O	O =1
R	N LEN(1) . X REM . N	:F(RETURN)
	X =EQ(X) 1
	O =O * X	:(R)

Try it online!

A function that takes input as a string.

answered Jan 16, 2018 at 21:12

Giuseppe

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gggg · Accepted Answer · 2018-01-16 23:31:59Z

1

Julia 0.6, 33 bytes

x->prod(filter(x->x>0,digits(x)))

Try it online!

answered Jan 16, 2018 at 23:31

gggg

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Dave · Accepted Answer · 2018-01-17 00:09:33Z

1

Pyth, 7 bytes

*FsMs#Q

Try it online!

*FsMs#Q Full program, takes input in "" from stdin and prints to stdout
    s#Q Filter input for truthiness of int-parse(digit), then
  sM    map int-parse to this,
*F      then fold over multiplication

answered Jan 17, 2018 at 0:09

Dave

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kakkarot · Accepted Answer · 2018-01-17 11:02:23Z

1

C# (.NET Core), 78 bytes

using System.Linq

n=>n.Replace("0","").Select(c=>c-'0').Aggregate(1,(a,b)=>a*b)

Try it online!

answered Jan 17, 2018 at 11:02

kakkarot

2791 silver badge4 bronze badges

\$\begingroup\$ Smaller is n=>n.Select(c=>c-'0').Aggregate(1,(a,b)=>a*(b==0?1:b)); You ought to include the terminating ';' \$\endgroup\$
– user230118
Jan 19, 2018 at 0:36
\$\begingroup\$ Following @user230118 suggestion, you can shrink it more by replacing '0' with 48, and (b==0?1:b) with (b>0?b:1) \$\endgroup\$
– auhmaan
Sep 23, 2019 at 16:50

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alephalpha · Accepted Answer · 2018-01-17 14:19:30Z

1

Pari/GP, 34 bytes

n->fold((a,b)->a*(b+!b),digits(n))

Try it online!

answered Jan 17, 2018 at 14:19

alephalpha

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

1

Pyth, 6 bytes

*F #sM

Try it Online!

Explanation:

    sM   Get digits of implicit string input
   #     Filter on identity, removing 0s
*F       Fold on multiplication

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 17, 2018 at 19:22

Steven H.

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DLosc · Accepted Answer · 2018-01-17 20:05:47Z

1

Pip, 7 bytes

$*YaDC0

Try it online!

Delete Character 0 from the argument a and fold ($) on multiplication (*). The Yank operator is used to adjust the precedence of the subexpression (one byte shorter than $*(aDC0)).

answered Jan 17, 2018 at 20:05

DLosc

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chromaticiT · Accepted Answer · 2018-01-19 00:47:49Z

1

Jelly, 4 bytes

Dḟ0P

Try it online!

Dḟ0P    main link

D       int to dec
 ḟ0     filter out 0
   P    product of list

answered Jan 19, 2018 at 0:47

chromaticiT

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MercyBeaucou · Accepted Answer · 2018-02-16 19:49:43Z

1

Befunge-98 (PyFunge), 14 bytes

1<*+!:-0'~.j@#

Try it online!

answered Feb 16, 2018 at 19:49

MercyBeaucou

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Jarom C · Accepted Answer · 2018-02-22 05:20:46Z

1

Microsoft Excel - 62 47 Bytes

{=PRODUCT(IFERROR(1/(1/MID(A1,ROW(A:A),1)),1))}

Explanation

Converts value in cell A1 to array of digits, changes 0s to 1s using absolute value, then multiplies.

15 bytes saved thanks to Engineer Toast

edited Feb 22, 2018 at 5:20

answered Jan 16, 2018 at 8:07

Jarom C

112 bronze badges

\$\begingroup\$ Save 15 bytes: {=PRODUCT(IFERROR(1/(1/MID(A1,ROW(A:A),1)),1))}. It drops zeroes by trying to divide by them to get an error. You can drop the INDIRECT stuff, too, because dividing by blank text values will also give you an error. All the errors are replaced by 1. \$\endgroup\$
– Engineer Toast
Jan 18, 2018 at 13:11

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

1

Stax, 4 bytes^CP437

┤caü

5 bytes when unpacked,

E0-k*

Run and debug online!

Explanation

E         Convert number to digits
 0-       Remove all zeros
   k*     Reduce array with multiplication

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Feb 22, 2018 at 5:41

Weijun Zhou

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The Non-Zero Digital Product Challenge

Challenge

Method

Test Cases

69 Answers 69

Haskell, 36 bytes

Jelly, 6, 5 bytes

><>, 19 bytes

Add++, 12 bytes

Clean, 55 bytes

APL+WIN, 9 bytes

Python 2, 46 44 bytes

Forth (gforth), 62 bytes

APL (Dyalog), 11 7 bytes

Lua, 45 bytes

Pyth, 10 9 Bytes

Explanation

Japt, 5 bytes

Explanation

Perl 6, 19 bytes

Expanded:

Batch, 78 bytes

Husk, 4 bytes

dc, 40 bytes

Red, 68 bytes

Javascript, 58 bytes

K (oK), 10 7 bytes

SNOBOL4 (CSNOBOL4), 83 bytes

Julia 0.6, 33 bytes

Pyth, 7 bytes

C# (.NET Core), 78 bytes

Pari/GP, 34 bytes

Pyth, 6 bytes

Pip, 7 bytes

Jelly, 4 bytes

Befunge-98 (PyFunge), 14 bytes

Microsoft Excel - 62 47 Bytes

Explanation

Stax, 4 bytesCP437

Explanation

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