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, showing that the single digit doesn't have to be 0:

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

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 2

C (gcc), 58 bytes

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

Try it online!

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;}

Try it online!

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

Answer 3

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);}

Try it online

Answer 4

V (vim), ^{43 30} 29 bytes

qqhYp:s/./*&/g
x0C<C-r>=<C-r>"
<esc>@qq@q

Try it online!

-13 bytes from DJMcMayhem. Go upvote their answer!

-1 byte from user41805.

Takes advantage of the fact that this only stops on single digit numbers.

Explanation

qqhYp:s/./*&/g
qq                  create a macro @q:
  l                  move one char right
                     this is the break condition, since you cant 
                     move to the right from a single digit
   Yp                duplicate the current line
     :s/./*&/g       add "*" before each digit

x0C<C-r>=<C-r>"     
x                    delete first char
 0C                  cut the current line and store in " register
   <C-r>=            evaluate the following string:
         <C-r>"      the value in " register
                     and paste in current line
 
@qq@q
@q                   recursively call macro
  q                 close macro
   @q               call macro q

Answer 5

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)

Try it online!

Answer 6

PicoLisp, 73 72 bytes

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

Try it online!

Answer 7

Japt, 15 bytes

I got some help from Shaggy on this one.

@pUÌì × Ì>9}f U

Run it online

Answer 8

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 9

Perl 6, 23 bytes

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

Try it online!

Answer 10

Husk, 5 bytes

U¡oΠd

Try it online!

Explanation

U¡oΠd
 ¡    iterate infinitely, collecting results of each iteration
  o   combination of 2 functions
    d get digits
   Π  multiply all of them
U     cut at first duplicate value

Answer 11

R, 59 55 bytes

Edit: -2 bytes thanks to Giuseppe, quickly superseded by -4 bytes (a different way) thanks to Robin Ryder

n=scan();while(print(n)>9)n=prod(utf8ToInt(c(n,""))-48)

Try it online!

A different method to Giuseppe's answer for the same number of bytes, here as a full recursive function instead of the (often-shorter) approach of taking input from the console using scan.

Answer 12

POSIX Shell + Utilities, 59 bytes

echo $1
[ ${#1} = 1 ]||$0 $(echo $1|fold -w1|paste -sd*|bc)

$ cat cg; echo; wc -c cg
echo $1
[ ${#1} = 1 ]||$0 $(echo $1|fold -w1|paste -sd*|bc)
59 cg
$ ./cg 277777788888899
277777788888899
4996238671872
438939648
4478976
338688
27648
2688
768
336
54
20
0
$ ./cg 25
25
10
0
$ ./cg 327
327
42
8
$ ./cg 68889
68889
27648
2688
768
336
54
20
0

Answer 13

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 14

ink, 66 bytes

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

Try it online!

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 15

Factor, 79 72 bytes

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

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Answer 16

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)

Try it online!

Answer 17

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 18

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 19

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 20

Attache, 28 27 bytes

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

Try it online!

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 21

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;}}}

Try it online!

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 22

Pyth, 9 bytes

j.u*FsM`N

Try it here!

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

Answer 23

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 24

Clojure, 73 bytes

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

Answer 25

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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Answer 26

Befunge-98 (FBBI), 43 42 bytes

&;1\v>#;:.:9`!#@_;
>$$  ^
^#::< '*%ap25/a_

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The first line takes input, prints the values and runs until a value <=9 is reached. The third line computes the product of digits of an integer (26 bytes on its own).

Answer 27

Pip, 9 bytes

W9<Pa$*:a

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Explanation

           a is command-line argument (implicit)
W          While...
   Pa      (print a's value)
 9<        ... a is greater than 9, i.e. more than one digit:
     $      Fold
        a   (the digits of) a
      *     on multiplication
       :    and assign back to a

Printing a in the loop header lets us output its value both before the loop and when the loop terminates.

Answer 28

Factor + math.unicode, 40 bytes

[ [ .s >dec 48 v-n Π dup 9 > ] loop . ]

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• [ ... dup 9 > ] loop loop while top of stack is greater than nine
• .s print stack non-destructively
• >dec convert number to string
• 48 v-n convert string to list of digits
• Π product of a list of numbers
• . print the leftover number (which will be the final output)

Answer 29

Python 3, 47 bytes

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

First Python 3 answer here.

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Answer 30

FunStack alpha, 38 bytes

Product Minus 48 Show compose3 fixiter

Try it at Replit: pass the input number as a command-line argument and enter the program on stdin.

Explanation

The fixiter modifier iterates a function until it reaches a fixed value, returning the full list of intermediate values. This is exactly what we want; there is a slight complication because the builtin ToBase function, which returns a list of digits, converts zero to an empty list rather than a list containing 0. So instead, we convert the number to a string and then convert each character to the corresponding digit.

                      compose3          Compose these three functions:
                 Show                     Convert number to string
        Minus 48                          Subtract 48 from each character
Product                                   Take the product of the character codes
                               fixiter  Iterate that function, stopping when a fixed point
                                        is reached, and return a list of intermediate values