Multiplicative persistence

Question

Multiplicative Persistence

1. Multiply all the digits in a number
2. Repeat until you have a single digit left

As explained by Numberphile:

• Numberphile "What's special about 277777788888899?"
• Numberphile "Multiplicative Persistence (extra footage)"

Example

1. 277777788888899 → 2x7x7x7x7x7x7x8x8x8x8x8x8x9x9 = 4996238671872
2. 4996238671872 → 4x9x9x6x2x3x8x6x7x1x8x7x2 = 438939648
3. 438939648 → 4x3x8x9x3x9x6x4x8 = 4478976
4. 4478976 → 4x4x7x8x9x7x6 = 338688
5. 338688 → 3x3x8x6x8x8 = 27648
6. 27648 → 2x7x6x4x8 = 2688
7. 2688 → 2x6x8x8 = 768
8. 768 → 7x6x8 = 336
9. 336 → 3x3x6 = 54
10. 54 → 5x4 = 20
11. 20 → 2x0 = 0

This is the current record, by the way: the smallest number with the largest number of steps.

Golf

A program that takes any whole number as input and then outputs the result of each step, starting with the input itself, until we hit a single digit. For 277777788888899 the output should be

277777788888899
4996238671872
438939648
4478976
338688
27648
2688
768
336
54
20
0

(Counting the number of steps is left as an exercise to the user).

More Examples

From A003001:

25
10
0

From A003001 as well:

68889
27648
2688
768
336
54
20
0

From the Numberphile video:

327
42
8

---

So there has been a question about Additive Persistence, but this is Multiplicative Persistence. Also, that question asks for the number of steps as output, while I'm interested in seeing the intermediate results.

Answer 1

Jelly, 4 bytes

DP$Ƭ

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Explanation

D    | convert to decimal digits
 P   | take the product
  $  | previous two links as a monad
   Ƭ | loop until no change, collecting all intermediate results

As a bonus, here's a TIO which will find the numbers with the largest number of steps for a given range of numbers of digits. It scales well even on TIO.

Answer 2

TI-BASIC (TI-84), 30 32 31 bytes

-1 byte thanks to @SolomonUcko!

While Ans>9:Disp Ans:prod(int(10fPart(Ans10^(seq(-X-1,X,0,log(Ans:End:Ans

Input is in Ans.
Output is displayed as the challenge requests. The trailing Ans is needed to print the last step.

I will admit, I did not think of this formula myself, rather I found it here and modified it to better fit the challenge.

EDIT: Upon rereading the challenge, I realized that the program must terminate if the product is one digit. Hence, 2 bytes were to be added to account for this.

Example:

24456756
        24456756
prgmCDGF8
        24456756
          201600
               0
11112
           11112
prgmCDGF8
           11112
               2

Explanation:

While Ans>9               ;loop until the product is one digit
Disp Ans                  ;display the current product
prod(                     ;get the product of...
 int(                     ; the integer part of...
  10fPart(                ; ten times the fractional part of...
  Ans                     ; each element in the following list times the
                          ;  current product
  10^(                    ; multiplied by the list generated by using each
                          ;  element of the following list as an exponent
                          ;  for 10^n
   seq(-X-1),X,0,log(Ans  ; generate a list of exponents from -1 to -L where
                          ;  L = the length of the current product
End
Ans                       ;leave the final product in "Ans" and implicitly
                          ; print it

Visual Model:
Ans starts off as 125673.
This model only covers the logic behind multiplying the digits; everything else is easier to understand.

seq(-X-1,X,0,log(Ans  =>  seq(-X-1,X,0,5.0992
   {-1 -2 -3 -4 -5 -6}
10^(...
   {.1 .01 .001 1E-4 1E-5 1E-6}
Ans...
   {12567.3 1256.73 125.673 12.5673 1.25673 .125673}
fPart(...
   {.3 .73 .673 .5673 .25673 .125673}
10...
   {3 7.3 6.73 5.673 2.5673 1.25673}
int(...
   {3 7 6 5 2 1}
   (the digits of the number, reversed)
prod(...
   1260
   (process is repeated again)

seq(-X-1,X,0,log(Ans  =>  seq(-X-1,X,0,3.1004
   {-1 -2 -3 -4}
10^(...
   {.1 .01 .001 1E-4}
Ans...
   {126 12.6 1.26 .126}
fPart(...
   {0 .6 .26 .126}
10...
   {0 6 2.6 1.26}
int(...
   {0 6 2 1}
prod(...
   0
   (product is less than 10.  loop ends)

---

Notes:

TI-BASIC is a tokenized language. Character count does not equal byte count.

10^( is this one-byte token.

This program will not provide the correct sequence of products with integers greater than 14 digits long due to the limitations of decimal precision on the TI calculators.

Answer 3

K (ngn/k), 9 bytes

{*/.'$x}\

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{ }\ keep applying the function in curly braces until the sequence converges

$x format the argument as a string (list of characters)

.' evaluate each (other dialects of k require a colon, .:')

*/ times over, i.e. product

Answer 4

05AB1E, 7 4 bytes

Δ=SP

Try it online or verify all test cases.

Explanation:

Δ     # Loop until the number no longer changes:
 =    #  Print the number with trailing newline (without popping the number itself)
      #  (which will be the implicit input in the first iteration)
  SP  #  Convert the number to a list of digits, and calculate its product

Answer 5

dzaima/APL, 14 11 bytes

∪{×/⍎¨⍕⍵}⍡≡

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Answer 6

R, 59 bytes

n=scan();while(print(n)>9)n=prod(n%/%10^(nchar(n):1-1)%%10)

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Since print invisibly returns its input, we can use print(n) inside the while loop to simulate a do-while loop. This is inspired by one of my tips for golfing in R.

The header helps prevent large numbers from being printed in scientific notation.

Answer 7

Wolfram Language (Mathematica), 47 bytes

Most@FixedPointList[Times@@IntegerDigits@#&,#]&

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Answer 8

Wolfram Language (Mathematica), 45 bytes

#//.x_/;x>9:>Times@@IntegerDigits@x&&Print@x&

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Answer 9

Perl 6, 23 bytes

{$_,{[*] .comb}…10>*}

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Answer 10

Python 2,  46  43 bytes

-3 thanks to xnor (chained comparison)

def f(n):print n;n>9>f(eval('*'.join(`n`)))

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Answer 11

Julia 0.7, 36 33 bytes

f(n)=n>9?[n;f(prod(digits(n)))]:n

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Thanks to H.PWiz for -3 bytes.

Answer 12

MathGolf, 9 10 bytes

h(ôo▒ε*h(→

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Now it correctly handles inputs that are single digits. Not perfect, but at least it is correct.

Explanation

h(            check length of input number and decrease by 1
  ö       →   while true with pop using the next 6 operators
   p          print with newline
    ▒         split to list of chars/digits
     ε*       reduce list by multiplication
       h(     length of TOS without popping, subtracted by 1 (exits when len(TOS) == 1)

Answer 13

C# (Visual C# Interactive Compiler), 79 74 68 bytes

void f(int a){Print(a);if(a>9)f((a+"").Aggregate(1,(j,k)=>k%48*j));}

I try to stay away from recursion in C# due to how long the method declaration is, but in this case it saves compared to a loop.

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Answer 14

Python 2, 61 62 59 bytes

def f(n):print n;n>9and f(reduce(int.__mul__,map(int,`n`)))

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_{-3 bytes, thanks to Jonathan Allan}

Answer 15

PowerShell, 54 bytes

for($a=$args;$a-gt9){$a;$a=("$a"|% t*y)-join"*"|iex}$a

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---

Iterative method that first writes the input argument, then converts it into a string and pipes it into a character array. This array is joined by a single asterisks, and executed as a command with the invoke expression alias. Since this writes Starting number down to the last number greater than 0, (20, in the given test scenario), I add a final $a to the end to output.

Answer 16

PHP, 63 bytes

<?=$n=$argn;while($n>9)echo"
",$n=array_product(str_split($n));

Iterative version, call with php -nF input from STDIN.

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PHP, 72 71 bytes

function h($n){echo"$n
",($n=array_product(str_split($n)))>9?h($n):$n;}

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Recursive version, as function.

Input: 277777788888899

277777788888899
4996238671872
438939648
4478976
338688
27648
2688
768
336
54
20
0

Input: 23

23
6

Answer 17

perl 5 (-n -M5.01), 32 30 25 bytes

say$_=eval;s/\B/*/g&&redo

25 bytes

30 bytes

32 bytes

Answer 18

JavaScript (ES6), 45 bytes

Returns an array of integers.

f=n=>[n,...n>9?f(eval([...n+''].join`*`)):[]]

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Answer 19

Ruby, 38 35 34 bytes

f=->n{p(n)>9&&f[eval n.digits*?*]}

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1 byte saved by by G B.

Answer 20

PowerShell, 51 bytes

filter f{$_
if($_-gt9){("$_"|% t*y)-join'*'|iex|f}}

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Answer 21

APL(NARS), 19 chars, 38 bytes

{⍵≤9:⍵⋄∇×/⍎¨⍕⍵⊣⎕←⍵}

test:

   f←{⍵≤9:⍵⋄∇×/⍎¨⍕⍵⊣⎕←⍵}
   f 23     
23
6
   f 27648     
27648
2688
768
336
54
20
0

Answer 22

Haskell, 45 bytes

f n=n:[x|n>9,x<-f$product$read.pure<$>show n]

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Answer 23

J, 16 bytes

([:*/,.&.":)^:a:

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Answer 24

Japt -R, 9 bytes

Horribly inefficient - don't even try to run the first test case!

_ì ×}hN â

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_ì ×}hN â     :Implicit input of integer U
      N       :Starting with the array of inputs (i.e., [U])
     h        :Do the following U times, pushing the result to N each time
_             :Take the last element in N and pass it through the following function
 ì            :  Convert to digit array
   ×          :  Reduce by multiplication
    }         :End function
        â     :Deduplicate N
              :Implicitly join with newlines and output

Answer 25

JavaScript (Babel Node), 46 bytes

f=a=>a>9?[a,...f(eval([...a+''].join`*`))]:[a]

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---

JavaScript (Babel Node), 44 bytes

If the input can be taken as String

f=a=>a>9?[a,...f(''+eval([...a].join`*`))]:a

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Answer 26

Brachylog, 7 bytes

ẉ?Ḋ|ẹ×↰

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Explanation

ẉ          Write the input followed by a linebreak
 ?Ḋ        If the input is a single digit, then it's over
   |       Otherwise
    ẹ      Split the input into a list of digits
     ×     Multiply them together
      ↰    Recursive call with the result of the multiplication as input

Answer 27

PowerShell, 64 59 bytes

for($a="$args";9-lt$a){$a;$a="$(($a|% t*y)-join'*'|iex)"}$a

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Iterative method. Takes input and stores it into $a, then enters a for loop so long as the length of $a is two or more (i.e., it's bigger than 9). Inside the loop we output $a and then recalculate it by converting it toCharArray, joining it together with *, and then iex (short for Invoke-Expression and similar to eval). Once we're out of the loop, we have a single digit left to print, so we place $a onto the pipeline again.

-5 bytes thanks to KGlasier.

Answer 28

Charcoal, 13 bytes

θＷ⊖Ｌθ«≔ＩΠθθ⸿θ

Try it online! Link is to verbose version of code. Explanation:

θ

Print the input for the first time.

Ｗ⊖Ｌθ«

Repeat while the length of the input is not 1.

≔ＩΠθθ

Replace the input with its digital product cast to string.

⸿θ

Print the input on a new line.

Answer 29

Retina, 24 bytes

.+~(\`

.
$&$*
^
.+¶$$.(

Try it online! Explanation:

.+~(\`

Print the current value on its own line at the start of every loop until it stops changing and don't print the unchanged value twice. Evaluate the current value at the end of each loop.

.
$&$*

Add a * after each digit.

^
.+¶$$.(

Finish turning the input into an expression that evaluates to the digital product.

Just for the record, Retina can do this in one line (25 bytes):

.+"¶"<~[".+¶$.("|'*]'*L`.

Answer 30

Java, 112 106 bytes

"Lossy conversion" thanks Java for the extra ~15 bytes
-6 bytes thanks to @Kevin Cruijssen
Obligatory stream abuse answer

n->{for(long t=10;t>9;n=(n+"").chars().mapToLong(i->i).reduce(1,(x,y)->x*(y-48)))System.out.println(t=n);}

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Answer 31

C (gcc), 58 bytes

f(n,t){for(;n=printf("%d\n",t=n)>2;)for(;n*=t%10,t/=10;);}

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Iterative approach turns out to be 1 byte shorter.

f(n,t){
    for(;n=printf("%d\n",t=n)   //print and update current number
            >2;)                //until only one digit is printed
        for(;n*=t%10,t/=10;);   //n*= product of digits of t (step)
}

---

C (gcc), 61 59 bytes (recursive)

f(n){printf("%d\n",n)>2&&f(p(n));}p(n){n=n?n%10*p(n/10):1;}

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Recursion seems to be shorter than iteration for both print and step...

Answer 32

Swift, 115 bytes

func m(a:Int){
print("\(a)")
var b=1,c=a
while c>0{
b*=c%10
c/=10}
b>9 ? m(a:b):print("\(b)")}
m(a:277777788888899)

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Answer 33

PicoLisp, 73 72 bytes

(de f(x)(if(> 10 x)(list x)(cons x(f(apply *(mapcar format(chop x)))))))

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Answer 34

Japt, 15 bytes

I got some help from Shaggy on this one.

@pUÌì × Ì>9}f U

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Answer 35

Chevron, 100 bytes

>^__>>^n
->+11?^n<10
>^n
^d<<1
^i<<1
^m<^i>^n
->+4??^m=^__
^d<<^d*^m
^i<<^i+1
->-4
^n<<^d
->-10
><^n

This is a fairly new language of my own creation - prototype interpreter, documentation, and example programs can be found at https://github.com/superloach/chevron.

Explanation:

• >^__>>^n - take the input as a NUM, with empty prompt
• ->+11?^n<10 - if the number is under 10 (1 digit), skip to the end
• >^n - output the current number
• ^d<<1 - initialise the product to 1
• ^i<<1 - initialise the character index to 1
• ^m<^i>^n - get i'th character of the number
• ->+4??^m=^__ - jump out of loop if no character is found
• ^d<<^d*^m - multiply character into product
• ^i<<^i+1 - increment character index
• ->-4 - hop back to continue getting characters
• ^n<<^d - once loop is done, assign product as new number
• ->-10 - hop all the way to the beginning for the next iteration
• ><^n - exit, printing the final number

Answer 36

java 8, 105 104 bytes

-1 byte thanks to @Benjamin Urquhart, replacing '0' with 48

i->{String s="";for(;(s+=i+"\n")!=""&i>9;){long m=1;for(byte c:(""+i).getBytes())m*=c-48;i=m;}return s;}

TIO

Answer 37

Factor, 79 72 bytes

: f ( x -- ) [ 10 >base [ 48 - ] map product dup dup . 9 > ] loop drop ;

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Answer 38

ink, 66 bytes

=p(n)
{n}{n<10:->->}
~temp v=1
-(a)~v=v*n%10
~n=n/10
{n:->a}->p(v)

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Explanation

=p(n)       // entry point - a stitch called p, with one parameter.
{n}         // print the value of n
{n<10:->->} // if n is a single digit, divert to wherever we entered from
~temp v=1   // declate a variable v, where we keep track of the product, and set it to 1
-(a)        // a labeled gather - we can jump here later
~v=v*n%10   // multiply v by (n mod 10) - the last digit in n
~n=n/10     // divide n by 10 - removing the last digit
{n:->a}     // if n is nonzero, go back to a
->p(v)      // if we reach this, divert to the top, this time with n set to the value of v

Doesn't feel very golfed, but I have no idea how you'd improve it.

Answer 39

Python 2, 87 bytes

b=input()
while 1:
 a=str(b);print a;b=1
 if len(a)<2:break
 for d in list(a):b*=int(d)

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Answer 40

SNOBOL4 (CSNOBOL4), 100 bytes

	P =INPUT
O	OUTPUT =N =P	:(S)
S	P =GT(N,9) 1	:F(END)
D	N LEN(1) . D REM . N	:F(O)
	P =P * D	:(D)
END

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A nice round 100 bytes!

	P =INPUT			;* Store input to P
O	OUTPUT =N =P	:(S)		;* print P and set N to P
S	P =GT(N,9) 1	:F(END)		;* if N > 9, set P (the digit product) to 1
D	N LEN(1) . D REM . N	:F(O)	;* take the first digit of N as D and set remaining digits to N
					;* if N is an empty string, goto label O
	P =P * D	:(D)		;* calculate the digit product, then return to D
END

Answer 41

C, 138 115 bytes

long atol(char*);main(c,v)char**v;{long z=1,n=atol(v[1]);for(;printf("%ld\n",n),n>9;n=z,z=1)for(;z*=n%10,n/=10;);}

Pass the number as a first command line argument.

Thanks to attinat for help with saving some bytes.

Ungolfed and prettyified:

long atol(char *);
main(c, v) char **v; {
    long z = 1,n = atol(v[1]);
    for(; printf("%ld\n", n), n > 9; n = z, z = 1)
        for(; z *= n % 10, n /= 10;);
}

Answer 42

Pyth, 9 bytes

j.u*FsM`N

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If we're allowed to output as an array, the j at the start is unnecessary, giving 8 bytes.

Answer 43

Racket, 107 bytes

(define(f n)(if(> 10 n)(list n)(cons n(f(apply *(map(λ(x)(-(char->integer x)48))(string->list(~v n))))))))

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Answer 44

C# (.NET Core), 192 191 180 bytes

using C=System.Console;class A{static void Main(string[] a){long i,k;var b=a[0];for(C.WriteLine(b);(k=b.Length)>1;C.WriteLine(b=i.ToString())){i=1;for(int j=0;j<k;)i*=b[j++]-48;}}}

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Pretty much just naive looping. I feel like that's a bit of a silly way to do it but here we are. Had a weird issue early on where the input didn't match the output, because it was overflowing an int, so here we are with longs. I could upgrade it to BigInteger if I needed to, though that would cost a few bytes.

I swear I always end up with off by ones when I do loops here, and it annoys me to no end.

Ungolfed-ish version:

using C=System.Console;
class A{
    static void Main(string[] a){
        long i,k;
        var b=a[0];
        for(C.WriteLine(b);(k=b.Length)>1;C.WriteLine(b=i.ToString())){
            i=1;
            for(int j=0;j<k;)i*=b[j++]-48;
        }
    }
}

Works for all integers less than int.MAX_VALUE digits long, C# (.NET Core), 204 bytes

using C=System.Console;class A{static void Main(string[] a){System.Numerics.BigInteger i;int j,k;var b=a[0];for(C.WriteLine(b);(k=b.Length)>1;C.WriteLine(b=i.ToString())){i=1;for(j=0;j<k;)i*=b[j++]-48;}}}

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Answer 45

Attache, 28 27 bytes

{If[_>9,_'$@Prod@List@_,_]}

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Oddly enough, the recursive approach, rather than a tacit approach, seems to be shortest. (There is a bug in NestListWhile involving currying, but even with the bugfix it'd be 3 bytes longer.)

Explanation

{
    If[ _ > 9 ,
        ?? if this is true, return a list
        [
            ?? with the current number
            _,
            ?? followed by this function called on the digital product
            ...$[ Prod[Digits[_]] ]
        ],
        ?? otherwise, return _
        _
    ]
}

Alternative

32 bytes: NestListWhile<~Prod@List,{_>9}~>

Answer 46

Stax, 4 bytes

▒σ╛g

Run and debug it

Unpacked, it's pretty simple.

gu  use the rest of the program to generate values until a duplicate is encountered
    output implicitly
E:* calculate product of digits

Answer 47

Perl 6, 23 bytes

{$_,{[*] .comb}...10>*}

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Answer 48

c++, lambda function, 72 bytes

[](auto n){while(cout<<n<<endl&&n>9){auto t=n;for(n=1;n*=t%10,t/=10;);}}

try it online

Answer 49

Clojure, 73 bytes

#(loop[i %](prn i)(if(> i 9)(recur(apply *(for[c(str i)](-(int c)48))))))

Answer 50

Scala, 83 bytes

def^(n:Long):Long={println(n);if(n>9)^((n+"").map(x=>(x-48).toLong).product)else n}

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