Digital root/Multiplicative digital root

From Rosetta Code
Task
Digital root/Multiplicative digital root
You are encouraged to solve this task according to the task description, using any language you may know.

The multiplicative digital root (MDR) and multiplicative persistence (MP) of a number, , is calculated rather like the Digital root except digits are multiplied instead of being added:

  1. Set to and to .
  2. While has more than one digit:
    • Find a replacement as the multiplication of the digits of the current value of .
    • Increment .
  3. Return (= MP) and (= MDR)


Task
  • Tabulate the MP and MDR of the numbers 123321, 7739, 893, 899998
  • Tabulate MDR versus the first five numbers having that MDR, something like:
MDR: [n0..n4]
===  ========
  0: [0, 10, 20, 25, 30]
  1: [1, 11, 111, 1111, 11111]
  2: [2, 12, 21, 26, 34]
  3: [3, 13, 31, 113, 131]
  4: [4, 14, 22, 27, 39]
  5: [5, 15, 35, 51, 53]
  6: [6, 16, 23, 28, 32]
  7: [7, 17, 71, 117, 171]
  8: [8, 18, 24, 29, 36]
  9: [9, 19, 33, 91, 119]

Show all output on this page.


References



Ada[edit]

The solution uses the Package "Generic_Root" from the additive digital roots [[1]].

with Ada.Text_IO, Generic_Root;   use Generic_Root;
 
procedure Multiplicative_Root is
 
procedure Compute is new Compute_Root("*"); -- "*" for multiplicative roots
 
package TIO renames Ada.Text_IO;
package NIO is new TIO.Integer_IO(Number);
 
procedure Print_Numbers(Target_Root: Number; How_Many: Natural) is
Current: Number := 0;
Root, Pers: Number;
begin
for I in 1 .. How_Many loop
loop
Compute(Current, Root, Pers);
exit when Root = Target_Root;
Current := Current + 1;
end loop;
NIO.Put(Current, Width => 6);
if I < How_Many then
TIO.Put(",");
end if;
Current := Current + 1;
end loop;
end Print_Numbers;
 
Inputs: Number_Array := (123321, 7739, 893, 899998);
Root, Pers: Number;
begin
TIO.Put_Line(" Number MDR MP");
for I in Inputs'Range loop
Compute(Inputs(I), Root, Pers);
NIO.Put(Inputs(I), Width => 8);
NIO.Put(Root, Width => 6);
NIO.Put(Pers, Width => 6);
TIO.New_Line;
end loop;
TIO.New_Line;
 
TIO.Put_Line(" MDR first_five_numbers_with_that_MDR");
for I in 0 .. 9 loop
TIO.Put(" " & Integer'Image(I) & " ");
Print_Numbers(Target_Root => Number(I), How_Many => 5);
TIO.New_Line;
end loop;
end Multiplicative_Root;
Output:
  Number   MDR    MP
  123321     8     3
    7739     8     3
     893     2     3
  899998     0     2

 MDR    first_five_numbers_with_that_MDR
   0       0,    10,    20,    25,    30
   1       1,    11,   111,  1111, 11111
   2       2,    12,    21,    26,    34
   3       3,    13,    31,   113,   131
   4       4,    14,    22,    27,    39
   5       5,    15,    35,    51,    53
   6       6,    16,    23,    28,    32
   7       7,    17,    71,   117,   171
   8       8,    18,    24,    29,    36
   9       9,    19,    33,    91,   119

ALGOL 68[edit]

Works with: ALGOL 68G version Any - tested with release 2.8.win32
# Multiplicative Digital Roots                                                #
 
# structure to hold the results of calculating the digital root & persistence #
MODE DR = STRUCT( INT root, INT persistence );
 
# calculate the multiplicative digital root and persistence of a number #
PROC md root = ( INT number )DR:
BEGIN
 
# calculate the product of the digits of a number #
PROC digit product = ( INT number )INT:
BEGIN
 
INT result := 1;
INT rest := number;
 
WHILE
result TIMESAB ( rest MOD 10 );
rest OVERAB 10;
rest > 0
DO
SKIP
OD;
 
result
END; # digit product #
 
INT mp := 0;
INT mdr := ABS number;
 
WHILE mdr > 9
DO
mp +:= 1;
mdr := digit product( mdr )
OD;
 
( mdr, mp )
END; # md root #
 
# prints a number and its MDR and MP #
PROC print md root = ( INT number )VOID:
BEGIN
DR mdr = md root( number );
print( ( whole( number, -6 )
, ": MDR: ", whole( root OF mdr, 0 )
, ", MP: ", whole( persistence OF mdr, -2 )
, newline
)
)
END; # print md root #
 
# prints the first few numbers with each possible Multiplicative Digital #
# Root. The number of values to print is specified as a parameter #
PROC tabulate mdr = ( INT number of values )VOID:
BEGIN
 
[ 0 : 9, 1 : number of values ]INT mdr values;
[ 0 : 9 ]INT mdr counts;
mdr counts[ AT 1 ] := ( 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 );
 
# find the first few numbers with each possible mdr #
 
INT values found := 0;
INT required values := 10 * number of values;
 
FOR value FROM 0 WHILE values found < required values
DO
DR mdr = md root( value );
IF mdr counts[ root OF mdr ] < number of values
THEN
# need more values with this multiplicative digital root #
values found +:= 1;
mdr counts[ root OF mdr ] +:= 1;
mdr values[ root OF mdr, mdr counts[ root OF mdr ] ] := value
FI
OD;
 
# print the values #
 
print( ( "MDR: [n0..n" + whole( number of values - 1, 0 ) + "]", newline ) );
print( ( "=== ========", newline ) );
FOR mdr pos FROM 1 LWB mdr values TO 1 UPB mdr values
DO
STRING separator := ": [";
print( ( whole( mdr pos, -3 ) ) );
FOR val pos FROM 2 LWB mdr values TO 2 UPB mdr values
DO
print( ( separator + whole( mdr values[ mdr pos, val pos ], 0 ) ) );
separator := ", "
OD;
print( ( "]", newline ) )
OD
 
END; # tabulate mdr #
 
main:(
print md root( 123321 );
print md root( 7739 );
print md root( 893 );
print md root( 899998 );
tabulate mdr( 5 )
)
Output:
123321: MDR: 8, MP:  3
  7739: MDR: 8, MP:  3
   893: MDR: 2, MP:  3
899998: MDR: 0, MP:  2
MDR: [n0..n4]
===  ========
  0: [0, 10, 20, 25, 30]
  1: [1, 11, 111, 1111, 11111]
  2: [2, 12, 21, 26, 34]
  3: [3, 13, 31, 113, 131]
  4: [4, 14, 22, 27, 39]
  5: [5, 15, 35, 51, 53]
  6: [6, 16, 23, 28, 32]
  7: [7, 17, 71, 117, 171]
  8: [8, 18, 24, 29, 36]
  9: [9, 19, 33, 91, 119]

ALGOL W[edit]

begin
 % calculate the Multiplicative Digital Root (mdr) and Multiplicative Persistence (mp) of n %
procedure getMDR ( integer value n
 ; integer result mdr, mp
) ;
begin
mp  := 0;
mdr := abs n;
while mdr > 9 do begin
integer v;
v  := mdr;
mdr := 1;
while begin
mdr := mdr * ( v rem 10 );
v  := v div 10;
v > 0
end do begin end;
mp := mp + 1;
end while_mdr_gt_9 ;
end getMDR ;
 
 % task test cases %
write( " N MDR MP" );
for n := 123321, 7739, 893, 899998 do begin
integer mdr, mp;
getMDR( n, mdr, mp );
write( s_w := 1, i_w := 8, n, i_w := 3, mdr, i_w := 2, mp )
end for_n ;
 
begin % find the first 5 numbers with each possible MDR %
integer requiredMdrs;
requiredMdrs := 5;
begin
integer array firstFew ( 0 :: 9, 1 :: requiredMdrs );
integer array mdrFOund ( 0 :: 9 );
integer totalFound, requiredTotal, n;
for i := 0 until 9 do mdrFound( i ) := 0;
totalFound  := 0;
requiredTotal := 10 * requiredMdrs;
n  := -1;
while totalFound < requiredTotal do begin
integer mdr, mp;
n := n + 1;
getMDR( n, mdr, mp );
if mdrFound( mdr ) < requiredMdrs then begin
 % found another number with this MDR and haven't found enough yet %
totalFound  := totalFound + 1;
mdrFound( mdr )  := mdrFound( mdr ) + 1;
firstFew( mdr, mdrFound( mdr ) ) := n
end if_found_another_MDR
end while_totalFound_lt_requiredTotal ;
 % print the table of MDRs andnumbers %
write( "MDR: [n0..n4]" );
write( "=== ========" );
for v := 0 until 9 do begin
write( i_w := 3, s_w := 0, v, ": [" );
for foundPos := 1 until requiredMdrs do begin
if foundPos > 1 then writeon( s_w := 0, ", " );
writeon( i_w := 1, s_w := 0, firstFew( v, foundPos ) )
end for_foundPos ;
writeon( s_w := 0, "]" )
end for_v
end
end
 
end.
Output:
       N MDR MP
  123321   8  3
    7739   8  3
     893   2  3
  899998   0  2
MDR: [n0..n4]
===  ========
  0: [0, 10, 20, 25, 30]
  1: [1, 11, 111, 1111, 11111]
  2: [2, 12, 21, 26, 34]
  3: [3, 13, 31, 113, 131]
  4: [4, 14, 22, 27, 39]
  5: [5, 15, 35, 51, 53]
  6: [6, 16, 23, 28, 32]
  7: [7, 17, 71, 117, 171]
  8: [8, 18, 24, 29, 36]
  9: [9, 19, 33, 91, 119]

AWK[edit]

# Multiplicative Digital Roots
 
BEGIN {
 
printMdrAndMp( 123321 );
printMdrAndMp( 7739 );
printMdrAndMp( 893 );
printMdrAndMp( 899998 );
 
tabulateMdr( 5 );
 
} # BEGIN
 
function printMdrAndMp( n )
{
calculateMdrAndMp( n );
printf( "%6d: MDR: %d, MP: %2d\n", n, MDR, MP );
} # printMdrAndMp
 
function calculateMdrAndMp( n, mdrStr, digit )
{
 
MP = 0; # global Multiplicative Persistence
MDR = ( n < 0 ? -n : n ); # global Multiplicative Digital Root
 
while( MDR > 9 )
{
MP ++;
mdrStr = "" MDR;
MDR = 1;
for( digit = 1; digit <= length( mdrStr ); digit ++ )
{
MDR *= ( substr( mdrStr, digit, 1 ) * 1 );
} # for digit
} # while MDR > 9
 
} # calculateMdrAndMp
 
function tabulateMdr( n, rqdValues, valueCount, value, pos )
{
 
# generate a table of the first n numbers with each possible MDR
 
rqdValues = n * 10;
valueCount = 0;
 
for( value = 0; valueCount < rqdValues; value ++ )
{
calculateMdrAndMp( value );
if( mdrCount[ MDR ] < n )
{
# still need another value with this MDR
valueCount ++;
mdrCount[ MDR ] ++;
mdrValues[ MDR ":" mdrCount[ MDR ] ] = value;
} # if mdrCount[ MDR ] < n
} # for value
 
# print the table
 
printf( "MDR: [n0..n%d]\n", n - 1 );
printf( "=== ========\n" );
 
for( pos = 0; pos < 10; pos ++ )
{
printf( "%3d:", pos );
separator = " [";
for( value = 1; value <= n; value ++ )
{
printf( "%s%d", separator, mdrValues[ pos ":" value ] );
separator = ", "
} # for value
printf( "]\n" );
} # for pos
 
} # tabulateMdr
Output:
123321: MDR: 8, MP:  3
  7739: MDR: 8, MP:  3
   893: MDR: 2, MP:  3
899998: MDR: 0, MP:  2
MDR: [n0..n4]
===  ========
  0: [0, 10, 20, 25, 30]
  1: [1, 11, 111, 1111, 11111]
  2: [2, 12, 21, 26, 34]
  3: [3, 13, 31, 113, 131]
  4: [4, 14, 22, 27, 39]
  5: [5, 15, 35, 51, 53]
  6: [6, 16, 23, 28, 32]
  7: [7, 17, 71, 117, 171]
  8: [8, 18, 24, 29, 36]
  9: [9, 19, 33, 91, 119]

Bracmat[edit]

(
& ( MP/MDR
= prod L n
. ( prod
= d
. @(!arg:%@?d ?arg)&!d*prod$!arg
| 1
)
& !arg:?L
& whl
' ( @(!arg:? [>1)
& (prod$!arg:?arg) !L:?L
)
& !L:? [?n
& (!n+-1.!arg)
)
& ( test
= n
.  !arg:%?n ?arg
& out$(!n "\t:" MP/MDR$!n)
& test$!arg
|
)
& test$(123321 7739 893 899998)
& 0:?i
& 1:?collecting:?done
& whl
' ( !i+1:?i
& MP/MDR$!i:(?MP.?MDR)
& ( !done:?*(!MDR.)^((?.)+?)*?
| (!MDR.)^(!i.)*!collecting:?collecting
& (  !collecting:?A*(!MDR.)^(?is+[5)*?Z
& !A*!Z:?collecting
& (!MDR.)^!is*!done:?done
|
)
)
& !collecting:~1
)
& whl
' ( !done:(?MDR.)^?is*?done
& put$(!MDR ":")
& whl'(!is:(?i.)+?is&put$(!i " "))
& put$\n
)
);
Output:
123321  : (3.8)
7739    : (3.8)
893     : (3.2)
899998  : (2.0)
0 :10  20  25  30  40
1 :1  11  111  1111  11111
2 :2  12  21  26  34
3 :3  13  31  113  131
4 :4  14  22  27  39
5 :5  15  35  51  53
6 :6  16  23  28  32
7 :7  17  71  117  171
8 :8  18  24  29  36
9 :9  19  33  91  119

C[edit]

 
#include <stdio.h>
 
#define twidth 5
#define mdr(rmdr, rmp, n)\
do { *rmp = 0; _mdr(rmdr, rmp, n); } while (0)

 
void _mdr(int *rmdr, int *rmp, long long n)
{
/* Adjust r if 0 case, so we don't return 1 */
int r = n ? 1 : 0;
while (n) {
r *= (n % 10);
n /= 10;
}
 
(*rmp)++;
if (r >= 10)
_mdr(rmdr, rmp, r);
else
*rmdr = r;
}
 
int main(void)
{
int i, j, vmdr, vmp;
const int values[] = { 123321, 7739, 893, 899998 };
const int vsize = sizeof(values) / sizeof(values[0]);
 
/* Initial test values */
printf("Number MDR MP\n");
for (i = 0; i < vsize; ++i) {
mdr(&vmdr, &vmp, values[i]);
printf("%6d  %3d  %3d\n", values[i], vmdr, vmp);
}
 
/* Determine table values */
int table[10][twidth] = { 0 };
int tfill[10] = { 0 };
int total = 0;
for (i = 0; total < 10 * twidth; ++i) {
mdr(&vmdr, &vmp, i);
if (tfill[vmdr] < twidth) {
table[vmdr][tfill[vmdr]++] = i;
total++;
}
}
 
/* Print calculated table values */
printf("\nMDR: [n0..n4]\n");
for (i = 0; i < 10; ++i) {
printf("%3d: [", i);
for (j = 0; j < twidth; ++j)
printf("%d%s", table[i][j], j != twidth - 1 ? ", " : "");
printf("]\n");
}
 
return 0;
}
 
Output:
Number    MDR    MP
123321     8     3
  7739     8     3
   893     2     3
899998     0     2

MDR: [n0..n4]
  0: [0, 10, 20, 25, 30]
  1: [1, 11, 111, 1111, 11111]
  2: [2, 12, 21, 26, 34]
  3: [3, 13, 31, 113, 131]
  4: [4, 14, 22, 27, 39]
  5: [5, 15, 35, 51, 53]
  6: [6, 16, 23, 28, 32]
  7: [7, 17, 71, 117, 171]
  8: [8, 18, 24, 29, 36]
  9: [9, 19, 33, 91, 119]

C#[edit]

using System;
using System.Collections.Generic;
using System.Linq;
 
class Program
{
static Tuple<int, int> DigitalRoot(long num)
{
int mp = 0;
while (num > 9)
{
num = num.ToString().ToCharArray().Select(x => x - '0').Aggregate((a, b) => a * b);
mp++;
}
return new Tuple<int, int>(mp, (int)num);
}
static void Main(string[] args)
{
foreach (long num in new long[] { 123321, 7739, 893, 899998 })
{
var t = DigitalRoot(num);
Console.WriteLine("{0} has multiplicative persistence {1} and multiplicative digital root {2}", num, t.Item1, t.Item2);
}
 
const int twidth = 5;
List<long>[] table = new List<long>[10];
for (int i = 0; i < 10; i++)
table[i] = new List<long>();
long number = -1;
while (table.Any(x => x.Count < twidth))
{
var t = DigitalRoot(++number);
if (table[t.Item2].Count < twidth)
table[t.Item2].Add(number);
}
for (int i = 0; i < 10; i++)
Console.WriteLine(" {0} : [{1}]", i, string.Join(", ", table[i]));
}
}
Output:
123321 has multiplicative persistence 3 and multiplicative digital root 8
7739 has multiplicative persistence 3 and multiplicative digital root 8
893 has multiplicative persistence 3 and multiplicative digital root 2
899998 has multiplicative persistence 2 and multiplicative digital root 0
 0 : [0, 10, 20, 25, 30]
 1 : [1, 11, 111, 1111, 11111]
 2 : [2, 12, 21, 26, 34]
 3 : [3, 13, 31, 113, 131]
 4 : [4, 14, 22, 27, 39]
 5 : [5, 15, 35, 51, 53]
 6 : [6, 16, 23, 28, 32]
 7 : [7, 17, 71, 117, 171]
 8 : [8, 18, 24, 29, 36]
 9 : [9, 19, 33, 91, 119]

C++[edit]

 
#include <iomanip>
#include <map>
#include <vector>
#include <iostream>
using namespace std;
 
void calcMDR( int n, int c, int& a, int& b )
{
int m = n % 10; n /= 10;
while( n )
{
m *= ( n % 10 );
n /= 10;
}
if( m >= 10 ) calcMDR( m, ++c, a, b );
else { a = m; b = c; }
}
 
void table()
{
map<int, vector<int> > mp;
int n = 0, a, b;
bool f = true;
while( f )
{
f = false;
calcMDR( n, 1, a, b );
mp[a].push_back( n );
n++;
for( int x = 0; x < 10; x++ )
if( mp[x].size() < 5 )
{ f = true; break; }
}
 
cout << "| MDR | [n0..n4]\n+-------+------------------------------------+\n";
for( int x = 0; x < 10; x++ )
{
cout << right << "| " << setw( 6 ) << x << "| ";
for( vector<int>::iterator i = mp[x].begin(); i != mp[x].begin() + 5; i++ )
cout << setw( 6 ) << *i << " ";
cout << "|\n";
}
cout << "+-------+------------------------------------+\n\n";
}
 
int main( int argc, char* argv[] )
{
cout << "| NUMBER | MDR | MP |\n+----------+----------+----------+\n";
int numbers[] = { 123321, 7739, 893, 899998 }, a, b;
for( int x = 0; x < 4; x++ )
{
cout << right << "| " << setw( 9 ) << numbers[x] << "| ";
calcMDR( numbers[x], 1, a, b );
cout << setw( 9 ) << a << "| " << setw( 9 ) << b << "|\n";
}
cout << "+----------+----------+----------+\n\n";
table();
return system( "pause" );
}
 
Output:
|  NUMBER  |   MDR    |    MP    |
+----------+----------+----------+
|    123321|         8|         3|
|      7739|         8|         3|
|       893|         2|         3|
|    899998|         0|         2|
+----------+----------+----------+

|  MDR  |  [n0..n4]
+-------+------------------------------------+
|      0|      0     10     20     25     30 |
|      1|      1     11    111   1111  11111 |
|      2|      2     12     21     26     34 |
|      3|      3     13     31    113    131 |
|      4|      4     14     22     27     39 |
|      5|      5     15     35     51     53 |
|      6|      6     16     23     28     32 |
|      7|      7     17     71    117    171 |
|      8|      8     18     24     29     36 |
|      9|      9     19     33     91    119 |
+-------+------------------------------------+

Common Lisp[edit]

 
(defun mdr/p (n)
"Return a list with MDR and MP of n"
(if (< n 10)
(list n 0)
(mdr/p-aux n 1 1)))
 
(defun mdr/p-aux (n a c)
(cond ((and (zerop n) (< a 10)) (list a c))
((zerop n) (mdr/p-aux a 1 (+ c 1)))
(t (mdr/p-aux (floor n 10) (* (rem n 10) a) c))))
 
(defun first-n-number-for-each-root (n &optional (r 0) (lst nil) (c 0))
"Return the first m number with MDR = 0 to 9"
(cond ((and (= (length lst) n) (= r 9)) (format t "[email protected]: ~a~%" r (reverse lst)))
((= (length lst) n) (format t "[email protected]: ~a~%" r (reverse lst))
(first-n-number-for-each-root n (+ r 1) nil 0))
((= (first (mdr/p c)) r) (first-n-number-for-each-root n r (cons c lst) (+ c 1)))
(t (first-n-number-for-each-root n r lst (+ c 1)))))
 
(defun start ()
(format t "Number: MDR MD~%")
(loop for el in '(123321 7739 893 899998)
do (format t "[email protected]: [email protected] ~}~%" el (mdr/p el)))
(format t "~%MDR: [n0..n4]~%")
(first-n-number-for-each-root 5))
Output:
Number: MDR  MD
123321:   8   3 
  7739:   8   3 
   893:   2   3 
899998:   0   2 

MDR: [n0..n4]
  0: (0 10 20 25 30)
  1: (1 11 111 1111 11111)
  2: (2 12 21 26 34)
  3: (3 13 31 113 131)
  4: (4 14 22 27 39)
  5: (5 15 35 51 53)
  6: (6 16 23 28 32)
  7: (7 17 71 117 171)
  8: (8 18 24 29 36)
  9: (9 19 33 91 119)

D[edit]

Translation of: Python
import std.stdio, std.algorithm, std.typecons, std.range, std.conv;
 
/// Multiplicative digital root.
auto mdRoot(in int n) pure /*nothrow*/ {
auto mdr = [n];
while (mdr.back > 9)
mdr ~= reduce!q{a * b}(1, mdr.back.text.map!(d => d - '0'));
//mdr ~= mdr.back.text.map!(d => d - '0').mul;
//mdr ~= mdr.back.reverseDigits.mul;
return tuple(mdr.length - 1, mdr.back);
}
 
void main() {
"Number: (MP, MDR)\n====== =========".writeln;
foreach (immutable n; [123321, 7739, 893, 899998])
writefln("%6d: (%s, %s)", n, n.mdRoot[]);
 
auto table = (int[]).init.repeat.enumerate!int.take(10).assocArray;
auto n = 0;
while (table.byValue.map!walkLength.reduce!min < 5) {
table[n.mdRoot[1]] ~= n;
n++;
}
"\nMP: [n0..n4]\n== ========".writeln;
foreach (const mp; table.byKey.array.sort())
writefln("%2d: %s", mp, table[mp].take(5));
}
Output:
Number: (MP, MDR)
======  =========
123321: (3, 8)
  7739: (3, 8)
   893: (3, 2)
899998: (2, 0)

MP: [n0..n4]
==  ========
 0: [0, 10, 20, 25, 30]
 1: [1, 11, 111, 1111, 11111]
 2: [2, 12, 21, 26, 34]
 3: [3, 13, 31, 113, 131]
 4: [4, 14, 22, 27, 39]
 5: [5, 15, 35, 51, 53]
 6: [6, 16, 23, 28, 32]
 7: [7, 17, 71, 117, 171]
 8: [8, 18, 24, 29, 36]
 9: [9, 19, 33, 91, 119]

Alternative Version[edit]

import std.stdio, std.algorithm, std.typecons, std.range;
 
uint digitsProduct(uint n) pure nothrow @nogc {
typeof(return) result = !!n;
while (n) {
result *= n % 10;
n /= 10;
}
return result;
}
 
/// Multiplicative digital root.
Tuple!(size_t, uint) mdRoot(uint m) pure nothrow {
auto mdr = m
.recurrence!((a, n) => a[n - 1].digitsProduct)
.until!q{ a <= 9 }(OpenRight.no).array;
return tuple(mdr.length - 1, mdr.back);
}
 
void main() {
"Number: (MP, MDR)\n====== =========".writeln;
foreach (immutable n; [123321, 7739, 893, 899998])
writefln("%6d: (%s, %s)", n, n.mdRoot[]);
 
auto table = (int[]).init.repeat.enumerate!int.take(10).assocArray;
auto n = 0;
while (table.byValue.map!walkLength.reduce!min < 5) {
table[n.mdRoot[1]] ~= n;
n++;
}
"\nMP: [n0..n4]\n== ========".writeln;
foreach (const mp; table.byKey.array.sort())
writefln("%2d: %s", mp, table[mp].take(5));
}

More Efficient Version[edit]

import std.stdio, std.algorithm, std.range;
 
/// Multiplicative digital root.
uint[2] mdRoot(in uint n) pure nothrow @nogc {
uint mdr = n;
uint count = 0;
 
while (mdr > 9) {
uint m = mdr;
uint digitsMul = !!m;
while (m) {
digitsMul *= m % 10;
m /= 10;
}
mdr = digitsMul;
count++;
}
 
return [count, mdr];
}
 
void main() {
"Number: [MP, MDR]\n====== =========".writeln;
foreach (immutable n; [123321, 7739, 893, 899998])
writefln("%6d: %s", n, n.mdRoot);
 
auto table = (int[]).init.repeat.enumerate!int.take(10).assocArray;
auto n = 0;
while (table.byValue.map!walkLength.reduce!min < 5) {
table[n.mdRoot[1]] ~= n;
n++;
}
"\nMP: [n0..n4]\n== ========".writeln;
foreach (const mp; table.byKey.array.sort())
writefln("%2d: %s", mp, table[mp].take(5));
}

The output is similar.

Elixir[edit]

defmodule Digital do
def mdroot(n), do: mdroot(n, 0)
 
defp mdroot(n, persist) when n < 10, do: {n, persist}
defp mdroot(n, persist), do: mdroot(product(n, 1), persist+1)
 
defp product(0, prod), do: prod
defp product(n, prod), do: product(div(n, 10), prod*rem(n, 10))
 
def task1(data) do
IO.puts "Number: MDR MP\n====== === =="
Enum.each(data, fn n ->
{mdr, persist} = mdroot(n)
 :io.format "~6w: ~w ~2w~n", [n, mdr, persist]
end)
end
 
def task2(m \\ 5) do
IO.puts "\nMDR: [n0..n#{m-1}]\n=== ========"
map = add_map(0, m, Map.new)
Enum.each(0..9, fn i ->
first = map[i] |> Enum.reverse |> Enum.take(m)
IO.puts " #{i}: #{inspect first}"
end)
end
 
defp add_map(n, m, map) do
{mdr, _persist} = mdroot(n)
new_map = Map.update(map, mdr, [n], fn vals -> [n | vals] end)
min_len = Map.values(new_map) |> Enum.map(&length(&1)) |> Enum.min
if min_len < m, do: add_map(n+1, m, new_map),
else: new_map
end
end
 
Digital.task1([123321, 7739, 893, 899998])
Digital.task2
Output:
Number: MDR  MP
======  ===  ==
123321:   8   3
  7739:   8   3
   893:   2   3
899998:   0   2

MDR: [n0..n4]
===  ========
  0: [0, 10, 20, 25, 30]
  1: [1, 11, 111, 1111, 11111]
  2: [2, 12, 21, 26, 34]
  3: [3, 13, 31, 113, 131]
  4: [4, 14, 22, 27, 39]
  5: [5, 15, 35, 51, 53]
  6: [6, 16, 23, 28, 32]
  7: [7, 17, 71, 117, 171]
  8: [8, 18, 24, 29, 36]
  9: [9, 19, 33, 91, 119]

Fortran[edit]

 
!Implemented by Anant Dixit (Oct, 2014)
program mdr
implicit none
integer :: i, mdr, mp, n, j
character(len=*), parameter :: hfmt = '(A18)', nfmt = '(I6)'
character(len=*), parameter :: cfmt = '(A3)', rfmt = '(I3)', ffmt = '(I9)'
 
write(*,hfmt) 'Number MDR MP '
write(*,*) '------------------'
 
i = 123321
call root_pers(i,mdr,mp)
write(*,nfmt,advance='no') i
write(*,cfmt,advance='no') ' '
write(*,rfmt,advance='no') mdr
write(*,cfmt,advance='no') ' '
write(*,rfmt) mp
 
i = 3939
call root_pers(i,mdr,mp)
write(*,nfmt,advance='no') i
write(*,cfmt,advance='no') ' '
write(*,rfmt,advance='no') mdr
write(*,cfmt,advance='no') ' '
write(*,rfmt) mp
 
i = 8822
call root_pers(i,mdr,mp)
write(*,nfmt,advance='no') i
write(*,cfmt,advance='no') ' '
write(*,rfmt,advance='no') mdr
write(*,cfmt,advance='no') ' '
write(*,rfmt) mp
 
i = 39398
call root_pers(i,mdr,mp)
write(*,nfmt,advance='no') i
write(*,cfmt,advance='no') ' '
write(*,rfmt,advance='no') mdr
write(*,cfmt,advance='no') ' '
write(*,rfmt) mp
 
write(*,*)
write(*,*)
write(*,*) 'First five numbers with MDR in first column: '
write(*,*) '---------------------------------------------'
 
do i = 0,9
n = 0
j = 0
write(*,rfmt,advance='no') i
do
call root_pers(j,mdr,mp)
if(mdr.eq.i) then
n = n+1
if(n.eq.5) then
write(*,ffmt) j
exit
else
write(*,ffmt,advance='no') j
end if
end if
j = j+1
end do
end do
 
end program
 
subroutine root_pers(i,mdr,mp)
implicit none
integer :: N, s, a, i, mdr, mp
n = i
a = 0
if(n.lt.10) then
mdr = n
mp = 0
return
end if
do while(n.ge.10)
a = a + 1
s = 1
do while(n.gt.0)
s = s * mod(n,10)
n = int(real(n)/10.0D0)
end do
n = s
end do
mdr = s
mp = a
end subroutine
 
 
Number   MDR   MP 
 ------------------
123321     8     3
  3939     2     4
  8822     0     3
 39398     0     3


 First five numbers with MDR in first column: 
 ---------------------------------------------
  0        0       10       20       25       30
  1        1       11      111     1111    11111
  2        2       12       21       26       34
  3        3       13       31      113      131
  4        4       14       22       27       39
  5        5       15       35       51       53
  6        6       16       23       28       32
  7        7       17       71      117      171
  8        8       18       24       29       36
  9        9       19       33       91      119

FreeBASIC[edit]

' FB 1.05.0 Win64
 
Function multDigitalRoot(n As UInteger, ByRef mp As Integer, base_ As Integer = 10) As Integer
Dim mdr As Integer
mp = 0
Do
mdr = IIf(n > 0, 1, 0)
While n > 0
mdr *= n Mod base_
n = n \ base_
Wend
mp += 1
n = mdr
Loop until mdr < base_
Return mdr
End Function
 
Dim As Integer mdr, mp
Dim a(3) As UInteger = {123321, 7739, 893, 899998}
For i As UInteger = 0 To 3
mp = 0
mdr = multDigitalRoot(a(i), mp)
Print a(i); Tab(10); "MDR ="; mdr; Tab(20); "MP ="; mp
Print
Next
Print
Print "MDR 1 2 3 4 5"
Print "=== ==========================="
Print
Dim num(0 To 9, 0 To 5) As UInteger '' all zero by default
Dim As UInteger n = 0, count = 0
Do
mdr = multDigitalRoot(n, mp)
If num(mdr, 0) < 5 Then
num(mdr, 0) += 1
num(mdr, num(mdr, 0)) = n
count += 1
End If
n += 1
Loop Until count = 50
 
For i As UInteger = 0 To 9
Print i; ":" ;
For j As UInteger = 1 To 5
Print Using "######"; num(i, j);
Next j
Print
Next i
 
Print
Print "Press any key to quit"
Sleep
Output:
123321   MDR = 8   MP = 3

7739     MDR = 8   MP = 3

893      MDR = 2   MP = 3

899998   MDR = 0   MP = 2


MDR    1     2     3     4     5
===  ===========================

0:     0    10    20    25    30
1:     1    11   111  1111 11111
2:     2    12    21    26    34
3:     3    13    31   113   131
4:     4    14    22    27    39
5:     5    15    35    51    53
6:     6    16    23    28    32
7:     7    17    71   117   171
8:     8    18    24    29    36
9:     9    19    33    91   119

Go[edit]

package main
 
import "fmt"
 
// Only valid for n > 0 && base >= 2
func mult(n uint64, base int) (mult uint64) {
for mult = 1; mult > 0 && n > 0; n /= uint64(base) {
mult *= n % uint64(base)
}
return
}
 
// Only valid for n >= 0 && base >= 2
func MultDigitalRoot(n uint64, base int) (mp, mdr int) {
var m uint64
for m = n; m >= uint64(base); mp++ {
m = mult(m, base)
}
return mp, int(m)
}
 
func main() {
const base = 10
const size = 5
 
const testFmt = "%20v %3v %3v\n"
fmt.Printf(testFmt, "Number", "MDR", "MP")
for _, n := range [...]uint64{
123321, 7739, 893, 899998,
18446743999999999999,
// From http://mathworld.wolfram.com/MultiplicativePersistence.html
3778888999, 277777788888899,
} {
mp, mdr := MultDigitalRoot(n, base)
fmt.Printf(testFmt, n, mdr, mp)
}
fmt.Println()
 
var list [base][]uint64
for i := range list {
list[i] = make([]uint64, 0, size)
}
for cnt, n := size*base, uint64(0); cnt > 0; n++ {
_, mdr := MultDigitalRoot(n, base)
if len(list[mdr]) < size {
list[mdr] = append(list[mdr], n)
cnt--
}
}
const tableFmt = "%3v: %v\n"
fmt.Printf(tableFmt, "MDR", "First")
for i, l := range list {
fmt.Printf(tableFmt, i, l)
}
}
Output:
              Number MDR  MP
              123321   8   3
                7739   8   3
                 893   2   3
              899998   0   2
18446743999999999999   0   2
          3778888999   0  10
     277777788888899   0  11

MDR: First
  0: [0 10 20 25 30]
  1: [1 11 111 1111 11111]
  2: [2 12 21 26 34]
  3: [3 13 31 113 131]
  4: [4 14 22 27 39]
  5: [5 15 35 51 53]
  6: [6 16 23 28 32]
  7: [7 17 71 117 171]
  8: [8 18 24 29 36]
  9: [9 19 33 91 119]

Haskell[edit]

Note that in the function mdrNums we don't know in advance how many numbers we'll need to examine to find the first 5 associated with all the MDRs. Using a lazy array to accumulate these numbers allows us to keep the function simple.

import Control.Arrow
import Data.Array
import Data.LazyArray
import Data.List (unfoldr)
import Data.Tuple
import Text.Printf
 
-- The multiplicative persistence (MP) and multiplicative digital root (MDR) of
-- the argument.
mpmdr :: Integer -> (Int, Integer)
mpmdr = (length *** head) . span (> 9) . iterate (product . digits)
 
-- Pairs (mdr, ns) where mdr is a multiplicative digital root and ns are the
-- first k numbers having that root.
mdrNums :: Int -> [(Integer, [Integer])]
mdrNums k = assocs $ lArrayMap (take k) (0,9) [(snd $ mpmdr n, n) | n <- [0..]]
 
digits :: Integral t => t -> [t]
digits 0 = [0]
digits n = unfoldr step n
where step 0 = Nothing
step k = Just (swap $ quotRem k 10)
 
printMpMdrs :: [Integer] -> IO ()
printMpMdrs ns = do
putStrLn "Number MP MDR"
putStrLn "====== == ==="
sequence_ [printf "%6d %2d %2d\n" n p r | n <- ns, let (p,r) = mpmdr n]
 
printMdrNums:: Int -> IO ()
printMdrNums k = do
putStrLn "MDR Numbers"
putStrLn "=== ======="
let showNums = unwords . map show
sequence_ [printf "%2d  %s\n" mdr $ showNums ns | (mdr,ns) <- mdrNums k]
 
main :: IO ()
main = do
printMpMdrs [123321, 7739, 893, 899998]
putStrLn ""
printMdrNums 5
Output:

Note that the values in the first column of the table are MDRs, as shown in the task's sample output, not MP as incorrectly stated in the task statement and column header.

Number MP MDR
====== == ===
123321  3  8
  7739  3  8
   893  3  2
899998  2  0

MDR Numbers
=== =======
 0  0 10 20 25 30
 1  1 11 111 1111 11111
 2  2 12 21 26 34
 3  3 13 31 113 131
 4  4 14 22 27 39
 5  5 15 35 51 53
 6  6 16 23 28 32
 7  7 17 71 117 171
 8  8 18 24 29 36
 9  9 19 33 91 119

Icon and Unicon[edit]

Works in both languages:

procedure main(A)
write(right("n",8)," ",right("MP",8),right("MDR",5))
every r := mdr(n := 123321|7739|893|899998) do
write(right(n,8),":",right(r[1],8),right(r[2],5))
write()
write(right("MDR",5)," ","[n0..n4]")
every m := 0 to 9 do {
writes(right(m,5),": [")
every writes(right((m = mdr(n := seq(m))[2],.n)\5,6))
write("]")
}
end
 
procedure mdr(m)
i := 0
while (.m > 10, m := multd(m), i+:=1)
return [i,m]
end
 
procedure multd(m)
c := 1
while m > 0 do c *:= 1(m%10, m/:=10)
return c
end
Output:
->drmdr
       n       MP  MDR
  123321:       3    8
    7739:       3    8
     893:       3    2
  899998:       2    0

  MDR  [n0..n4]
    0: [     0    20    30    40    45]
    1: [     1    11   111  1111 11111]
    2: [     2    12    21    26    34]
    3: [     3    13    31   113   131]
    4: [     4    14    22    27    39]
    5: [     5    15    35    51    53]
    6: [     6    16    23    28    32]
    7: [     7    17    71   117   171]
    8: [     8    18    24    29    36]
    9: [     9    19    33    91   119]
->

J[edit]

First, we need something to split a number into digits:

   10&#.inv 123321
1 2 3 3 2 1

Second, we need to find their product:

   */@(10&#.inv) 123321
36

Then we use this inductively until it converges:

   */@(10&#.inv)^:a: 123321
123321 36 18 8

MP is one less than the length of this list, and MDR is the last element of this list:

   (<:@#,{:) */@(10&#.inv)^:a: 123321
3 8
(<:@#,{:) */@(10&#.inv)^:a: 7739
3 8
(<:@#,{:) */@(10&#.inv)^:a: 893
3 2
(<:@#,{:) */@(10&#.inv)^:a: 899998
2 0

For the table, we don't need that whole list, we only need the final value. Then use these values to classify the original argument (taking the first five from each group):

   (5&{./.~ (*/@(10&#.inv)^:_)"0) i.20000
0 10 20 25 30
1 11 111 1111 11111
2 12 21 26 34
3 13 31 113 131
4 14 22 27 39
5 15 35 51 53
6 16 23 28 32
7 17 71 117 171
8 18 24 29 36
9 19 33 91 119

Note that since the first 10 non-negative integers are single digit values, the first column here doubles as a label (representing the corresponding multiplicative digital root).

Java[edit]

Works with: Java version 8
import java.util.*;
 
public class MultiplicativeDigitalRoot {
 
public static void main(String[] args) {
 
System.out.println("NUMBER MDR MP");
for (long n : new long[]{123321, 7739, 893, 899998}) {
long[] a = multiplicativeDigitalRoot(n);
System.out.printf("%6d %4d %4d%n", a[0], a[1], a[2]);
}
 
System.out.println();
 
Map<Long, List<Long>> table = new HashMap<>();
for (long i = 0; i < 10; i++)
table.put(i, new ArrayList<>());
 
for (long cnt = 0, n = 0; cnt < 10;) {
long[] res = multiplicativeDigitalRoot(n++);
List<Long> list = table.get(res[1]);
if (list.size() < 5) {
list.add(res[0]);
cnt = list.size() == 5 ? cnt + 1 : cnt;
}
}
 
System.out.println("MDR: first five numbers with same MDR");
table.forEach((key, lst) -> {
System.out.printf("%3d: ", key);
lst.forEach(e -> System.out.printf("%6s ", e));
System.out.println();
});
}
 
public static long[] multiplicativeDigitalRoot(long n) {
int mp = 0;
long mdr = n;
while (mdr > 9) {
long m = mdr;
long total = 1;
while (m > 0) {
total *= m % 10;
m /= 10;
}
mdr = total;
mp++;
}
return new long[]{n, mdr, mp};
}
}
NUMBER  MDR   MP
123321    8    3
  7739    8    3
   893    2    3
899998    0    2

MDR: first five numbers with same MDR
  0:      0     10     20     25     30 
  1:      1     11    111   1111  11111 
  2:      2     12     21     26     34 
  3:      3     13     31    113    131 
  4:      4     14     22     27     39 
  5:      5     15     35     51     53 
  6:      6     16     23     28     32 
  7:      7     17     71    117    171 
  8:      8     18     24     29     36 
  9:      9     19     33     91    119 

jq[edit]

def do_until(condition; next):
def u: if condition then . else (next|u) end;
u;
 
def mdroot(n):
def multiply: reduce .[] as $i (1; .*$i);
# state: [mdr, persist]
[n, 0]
| do_until( .[0] < 10;
[(.[0] | tostring | explode | map(.-48) | multiply), .[1] + 1]
);
 
# Produce a table with 10 rows (numbered from 0),
# showing the first n numbers having the row-number as the mdr
def tabulate(n):
# state: [answer_matrix, next_i]
def tab:
def minlength: map(length) | min;
.[0] as $matrix
| .[1] as $i
| if (.[0]|minlength) == n then .[0]
else (mdroot($i) | .[0]) as $mdr
| if $matrix[$mdr]|length < n then
($matrix[$mdr] + [$i]) as $row
| $matrix | setpath([$mdr]; $row)
else $matrix
end
| [ ., $i + 1 ]
| tab
end;
 
[[], 0] | tab;
Example:
 
def neatly:
. as $in
| range(0;length)
| "\(.): \($in[.])";
 
def rjust(n): tostring | (n-length)*" " + .;
 
# The task:
" i  : [MDR, MP]",
((123321, 7739, 893, 899998) as $i
| "\($i|rjust(6)): \(mdroot($i))"),
"",
"Tabulation",
"MDR: [n0..n4]",
(tabulate(5) | neatly)
Output:
$ jq -n -r -c -f mdr.jq
 
i  : [MDR, MP]
123321: [8,3]
7739: [8,3]
893: [2,3]
899998: [0,2]
 
Tabulation
MDR: [n0..n4]
0: [0,10,20,25,30]
1: [1,11,111,1111,11111]
2: [2,12,21,26,34]
3: [3,13,31,113,131]
4: [4,14,22,27,39]
5: [5,15,35,51,53]
6: [6,16,23,28,32]
7: [7,17,71,117,171]
8: [8,18,24,29,36]
9: [9,19,33,91,119]

Julia[edit]

Function

 
function digitalmultroot{S<:Integer,T<:Integer}(n::S, bs::T=10)
-1 < n && 1 < bs || throw(DomainError())
ds = n
pers = 0
while bs <= ds
ds = prod(digits(ds, bs))
pers += 1
end
return (pers, ds)
end
 

Main

 
const bs = 10
const excnt = 5
 
println("Testing Multiplicative Digital Root.\n")
for i in [123321, 7739, 893, 899998]
(pers, ds) = digitalmultroot(i, bs)
print(@sprintf("%8d", i))
print(" has persistence ", pers)
println(" and digital root ", ds)
end
 
dmr = zeros(Int, bs, excnt)
hasroom = trues(bs)
dex = ones(Int, bs)
 
i = 0
while any(hasroom)
(pers, ds) = digitalmultroot(i, bs)
ds += 1
if hasroom[ds]
dmr[ds, dex[ds]] = i
dex[ds] += 1
if dex[ds] > excnt
hasroom[ds] = false
end
end
i += 1
end
 
println("\n MDR: First ", excnt, " numbers having this MDR")
for (i, d) in enumerate(0:(bs-1))
print(@sprintf("%4d: ", d))
println(join([@sprintf("%6d", dmr[i, j]) for j in 1:excnt], ","))
end
 
Output:
Testing Multiplicative Digital Root.

  123321 has persistence 3 and digital root 8
    7739 has persistence 3 and digital root 8
     893 has persistence 3 and digital root 2
  899998 has persistence 2 and digital root 0

 MDR:    First 5 numbers having this MDR
   0:      0,    10,    20,    25,    30
   1:      1,    11,   111,  1111, 11111
   2:      2,    12,    21,    26,    34
   3:      3,    13,    31,   113,   131
   4:      4,    14,    22,    27,    39
   5:      5,    15,    35,    51,    53
   6:      6,    16,    23,    28,    32
   7:      7,    17,    71,   117,   171
   8:      8,    18,    24,    29,    36
   9:      9,    19,    33,    91,   119

Kotlin[edit]

Translation of: FreeBASIC
// version 1.1.2
 
fun multDigitalRoot(n: Int): Pair<Int, Int> = when {
n < 0 -> throw IllegalArgumentException("Negative numbers not allowed")
else -> {
var mdr: Int
var mp = 0
var nn = n
do {
mdr = if (nn > 0) 1 else 0
while (nn > 0) {
mdr *= nn % 10
nn /= 10
}
mp++
nn = mdr
}
while (mdr >= 10)
Pair(mdr, mp)
}
}
 
fun main(args: Array<String>) {
val ia = intArrayOf(123321, 7739, 893, 899998)
for (i in ia) {
val (mdr, mp) = multDigitalRoot(i)
println("${i.toString().padEnd(9)} MDR = $mdr MP = $mp")
}
println()
println("MDR n0 n1 n2 n3 n4")
println("=== ===========================")
val ia2 = Array(10) { IntArray(6) } // all zero by default
var n = 0
var count = 0
do {
val (mdr, _) = multDigitalRoot(n)
if (ia2[mdr][0] < 5) {
ia2[mdr][0]++
ia2[mdr][ia2[mdr][0]] = n
count++
}
n++
}
while (count < 50)
 
for (i in 0..9) {
print("$i:")
for (j in 1..5) print("%6d".format(ia2[i][j]))
println()
}
}
Output:
123321    MDR = 8  MP = 3
7739      MDR = 8  MP = 3
893       MDR = 2  MP = 3
899998    MDR = 0  MP = 2

MDR   n0    n1    n2    n3    n4
===  ===========================
0:     0    10    20    25    30
1:     1    11   111  1111 11111
2:     2    12    21    26    34
3:     3    13    31   113   131
4:     4    14    22    27    39
5:     5    15    35    51    53
6:     6    16    23    28    32
7:     7    17    71   117   171
8:     8    18    24    29    36
9:     9    19    33    91   119

Mathematica / Wolfram Language[edit]

 
ClearAll[mdr, mp, nums];
mdr[n_] := NestWhile[Times @@ IntegerDigits[#] &, n, # > 9 &];
mp[n_] := [email protected][Times @@ IntegerDigits[#] &, n, # > 9 &] - 1;
TableForm[{#, mdr[#], mp[#]} & /@ {123321, 7739, 893, 899998},
TableHeadings -> {None, {"Number", "MDR", "MP"}}]
nums = ConstantArray[{}, 10];
For[i = 0, Min[Length /@ nums] < 5, i++, AppendTo[nums[[mdr[i] + 1]], i]];
TableForm[Table[{i, Take[nums[[i + 1]], 5]}, {i, 0, 9}],
TableHeadings -> {None, {"MDR", "First 5"}}, TableDepth -> 2]
 
Output:
Number   MDR   MP
-----------------
123321   8     3
7739     8     3
893      2     3
899998   0     2

MDR   First 5
-----------------------------
0   {0, 10, 20, 25, 30}
1   {1, 11, 111, 1111, 11111}
2   {2, 12, 21, 26, 34}
3   {3, 13, 31, 113, 131}
4   {4, 14, 22, 27, 39}
5   {5, 15, 35, 51, 53}
6   {6, 16, 23, 28, 32}
7   {7, 17, 71, 117, 171}
8   {8, 18, 24, 29, 36}
9   {9, 19, 33, 91, 119}

Nim[edit]

Translation of: Python
import strutils, future
 
template newSeqWith(len: int, init: expr): expr =
var result {.gensym.} = newSeq[type(init)](len)
for i in 0 .. <len:
result[i] = init
result
 
proc mdroot(n): tuple[mp, mdr: int] =
var mdr = @[n]
while mdr[mdr.high] > 9:
var n = 1
for dig in $mdr[mdr.high]:
n *= parseInt($dig)
mdr.add n
(mdr.high, mdr[mdr.high])
 
for n in [123321, 7739, 893, 899998]:
echo align($n, 6)," ",mdroot(n)
echo ""
 
var table = newSeqWith(10, newSeq[int]())
for n in 0..int.high:
if table.map((x: seq[int]) => x.len).min >= 5: break
table[mdroot(n).mdr].add n
 
for mp, val in table:
echo mp,": ",val[0..4]
Output:
123321 (mp: 3, mdr: 8)
  7739 (mp: 3, mdr: 8)
   893 (mp: 3, mdr: 2)
899998 (mp: 2, mdr: 0)

0: @[0, 10, 20, 25, 30]
1: @[1, 11, 111, 1111, 11111]
2: @[2, 12, 21, 26, 34]
3: @[3, 13, 31, 113, 131]
4: @[4, 14, 22, 27, 39]
5: @[5, 15, 35, 51, 53]
6: @[6, 16, 23, 28, 32]
7: @[7, 17, 71, 117, 171]
8: @[8, 18, 24, 29, 36]
9: @[9, 19, 33, 91, 119]

PARI/GP[edit]

a(n)=my(i);while(n>9,n=factorback(digits(n));i++);[i,n];
apply(a, [123321, 7739, 893, 899998])
v=vector(10,i,[]); forstep(n=0,oo,1, t=a(n)[2]+1; if(#v[t]<5,v[t]=concat(v[t],n); if(vecmin(apply(length,v))>4, return(v))))
Output:
%1 = [[3, 8], [3, 8], [3, 2], [2, 0]]
%2 = [[0, 10, 20, 25, 30], [1, 11, 111, 1111, 11111], [2, 12, 21, 26, 34], [3, 13, 31, 113, 131], [4, 14, 22, 27, 39], [5, 15, 35, 51, 53], [6, 16, 23, 28, 32], [7, 17, 71, 117, 171], [8, 18, 24, 29, 36], [9, 19, 33, 91, 119]]

Perl[edit]

Translation of: D
use warnings;
use strict;
 
sub mdr {
my $n = shift;
my($count, $mdr) = (0, $n);
while ($mdr > 9) {
my($m, $dm) = ($mdr, 1);
while ($m) {
$dm *= $m % 10;
$m = int($m/10);
}
$mdr = $dm;
$count++;
}
($count, $mdr);
}
 
print "Number: (MP, MDR)\n====== =========\n";
foreach my $n (123321, 7739, 893, 899998) {
printf "%6d: (%d, %d)\n", $n, mdr($n);
}
print "\nMP: [n0..n4]\n== ========\n";
foreach my $target (0..9) {
my $i = 0;
my @n = map { $i++ while (mdr($i))[1] != $target; $i++; } 1..5;
print " $target: [", join(", ", @n), "]\n";
}
Output:
Number: (MP, MDR)
======  =========
123321: (3, 8)
  7739: (3, 8)
   893: (3, 2)
899998: (2, 0)

MP: [n0..n4]
==  ========
 0: [0, 10, 20, 25, 30]
 1: [1, 11, 111, 1111, 11111]
 2: [2, 12, 21, 26, 34]
 3: [3, 13, 31, 113, 131]
 4: [4, 14, 22, 27, 39]
 5: [5, 15, 35, 51, 53]
 6: [6, 16, 23, 28, 32]
 7: [7, 17, 71, 117, 171]
 8: [8, 18, 24, 29, 36]
 9: [9, 19, 33, 91, 119]

Perl 6[edit]

sub multiplicative-digital-root(Int $n) {
return .elems - 1, .[.end]
given cache($n, {[*] .comb} ... *.chars == 1)
}
 
for 123321, 7739, 893, 899998 {
say "$_: ", .&multiplicative-digital-root;
}
 
for ^10 -> $d {
say "$d : ", .[^5]
given (1..*).grep: *.&multiplicative-digital-root[1] == $d;
}
Output:
123321: 3 8
7739: 3 8
893: 3 2
899998: 2 0
0 : 10 20 25 30 40
1 : 1 11 111 1111 11111
2 : 2 12 21 26 34
3 : 3 13 31 113 131
4 : 4 14 22 27 39
5 : 5 15 35 51 53
6 : 6 16 23 28 32
7 : 7 17 71 117 171
8 : 8 18 24 29 36
9 : 9 19 33 91 119

PL/I[edit]

version 1[edit]

This example is incomplete. Missing second half of task! Please ensure that it meets all task requirements and remove this message.
multiple: procedure options (main);  /* 29 April 2014 */
 
declare n fixed binary (31);
 
find_mdr: procedure;
declare (mdr, mp, p) fixed binary (31);
 
mdr = n;
do mp = 1 by 1 until (p <= 9);
p = 1;
do until (mdr = 0); /* Form product of the digits in mdr. */
p = mod(mdr, 10) * p;
mdr= mdr/10;
end;
mdr = p;
end;
put skip data (n, mdr, mp);
end find_mdr;
 
do n = 123321, 7739, 893, 899998;
call find_mdr;
end;
 
end multiple;
Output:
N=        123321        MDR=             8      MP=             3;
N=          7739        MDR=             8      MP=             3;
N=           893        MDR=             2      MP=             3;
N=        899998        MDR=             0      MP=             2;

version 2[edit]

 mdrt: Proc Options(main);
Dcl (x,p,r) Bin Fixed(31);
Put Edit('number persistence multiplicative digital root')(Skip,a);
Put Edit('------- ----------- ---------------------------')(Skip,a);
Call task1(123321);
Call task1( 7739);
Call task1( 893);
Call task1(899998);
 
task1: Procedure(x);
Dcl x Bin Fixed(31);
Call mdr(x,p,r);
Put Edit(x,p,r)(Skip,f(8),f(8),f(22));
End;
 
Dcl zn(0:9) Bin Fixed(31);
Dcl z(0:9,5) Bin Fixed(31);
zn=0;
zn(0)=1;
z(0,1)=0;
Do x=1 To 11111;
Call mdr(x,p,r);
If zn(r)<5 Then Do;
zn(r)+=1;
z(r,zn(r))=x;
End;
End;
Put Edit(' ')(Skip,a);
Put Edit('MDR first 5 numbers that have a matching MDR')(Skip,a);
Put Edit('--- ----------------------------------------')(Skip,a);
 
Do r=0 To 9;
Put Edit(r,' ')(Skip,f(3),a);
Do i=1 To 5;
Put Edit(z(r,i))(f(6));
End;
End;
 
mdr: Procedure(y,p,r);
Dcl (y,p,r) Bin Fixed(31);
Dcl (k,yy) Bin Fixed(31);
Dcl pic Pic'(10)9';
Dcl d Pic'9';
pic=abs(y);
Do p=1 By 1 Until(pic<10);
Do k=1 To 10 Until(substr(pic,k,1)>'0');
End;
r=1;
Do k=k To 10;
d=substr(pic,k,1);
r=r*d;
End;
pic=r;
End;
End;
End;
Output:
number   persistence   multiplicative digital root
-------  -----------   ---------------------------
  123321       3                     8
    7739       3                     8
     893       3                     2
  899998       2                     0

MDR  first 5 numbers that have a matching MDR
---  ----------------------------------------
  0       0    10    20    25    30
  1       1    11   111  1111 11111
  2       2    12    21    26    34
  3       3    13    31   113   131
  4       4    14    22    27    39
  5       5    15    35    51    53
  6       6    16    23    28    32
  7       7    17    71   117   171
  8       8    18    24    29    36
  9       9    19    33    91   119

Python[edit]

Python: Inspired by the solution to the Digital root task[edit]

try:
from functools import reduce
except:
pass
 
def mdroot(n):
'Multiplicative digital root'
mdr = [n]
while mdr[-1] > 9:
mdr.append(reduce(int.__mul__, (int(dig) for dig in str(mdr[-1])), 1))
return len(mdr) - 1, mdr[-1]
 
if __name__ == '__main__':
print('Number: (MP, MDR)\n====== =========')
for n in (123321, 7739, 893, 899998):
print('%6i: %r' % (n, mdroot(n)))
 
table, n = {i: [] for i in range(10)}, 0
while min(len(row) for row in table.values()) < 5:
mpersistence, mdr = mdroot(n)
table[mdr].append(n)
n += 1
print('\nMP: [n0..n4]\n== ========')
for mp, val in sorted(table.items()):
print('%2i: %r' % (mp, val[:5]))
Output:
Number: (MP, MDR)
======  =========
123321: (3, 8)
  7739: (3, 8)
   893: (3, 2)
899998: (2, 0)

MP: [n0..n4]
==  ========
 0: [0, 10, 20, 25, 30]
 1: [1, 11, 111, 1111, 11111]
 2: [2, 12, 21, 26, 34]
 3: [3, 13, 31, 113, 131]
 4: [4, 14, 22, 27, 39]
 5: [5, 15, 35, 51, 53]
 6: [6, 16, 23, 28, 32]
 7: [7, 17, 71, 117, 171]
 8: [8, 18, 24, 29, 36]
 9: [9, 19, 33, 91, 119]

Python: Inspired by the more efficient version of D.[edit]

Substitute the following function to run twice as fast when calculating mdroot(n) with n in range(1000000).

def mdroot(n):
count, mdr = 0, n
while mdr > 9:
m, digitsMul = mdr, 1
while m:
m, md = divmod(m, 10)
digitsMul *= md
mdr = digitsMul
count += 1
return count, mdr
Output:

(Exactly the same as before).

Racket[edit]

#lang racket
(define (digital-product n)
(define (inr-d-p m rv)
(cond
[(zero? m) rv]
[else (define-values (q r) (quotient/remainder m 10))
(if (zero? r) 0 (inr-d-p q (* rv r)))])) ; lazy on zero
(inr-d-p n 1))
 
(define (mdr/mp n)
(define (inr-mdr/mp m i)
(if (< m 10) (values m i) (inr-mdr/mp (digital-product m) (add1 i))))
(inr-mdr/mp n 0))
 
(printf "Number\tMDR\tmp~%======\t===\t==~%")
(for ((n (in-list '(123321 7739 893 899998))))
(define-values (mdr mp) (mdr/mp n))
(printf "~a\t~a\t~a~%" n mdr mp))
 
(printf "~%MDR\t[n0..n4]~%===\t========~%")
(for ((MDR (in-range 10)))
(define (has-mdr? n) (define-values (mdr mp) (mdr/mp n)) (= mdr MDR))
(printf "~a\t~a~%" MDR (for/list ((_ 5) (n (sequence-filter has-mdr? (in-naturals)))) n)))
Output:
Number	MDR	mp
======	===	==
123321	8	3
7739	8	3
893	2	3
899998	0	2

MDR	[n0..n4]
===	========
0	(0 10 20 25 30)
1	(1 11 111 1111 11111)
2	(2 12 21 26 34)
3	(3 13 31 113 131)
4	(4 14 22 27 39)
5	(5 15 35 51 53)
6	(6 16 23 28 32)
7	(7 17 71 117 171)
8	(8 18 24 29 36)
9	(9 19 33 91 119)

REXX[edit]

idomatic version[edit]

/*REXX pgm finds persistence and multiplicative digital root of some #'s*/
numeric digits 100 /*increase the number of digits. */
parse arg x /*get some numbers from the C.L. */
if x='' then x=123321 7739 893 899998 /*use defaults if none specified.*/
say center('number',8) ' persistence multiplicative digital root'
say copies('─' ,8) ' ─────────── ───────────────────────────'
/* [↑] title and separator. */
do j=1 for words(x); n=word(x,j) /*process each number in the list*/
parse value mdr(n) with mp mdr /*obtain the persistence and MDR.*/
say right(n,8) center(mp,13) center(mdr,30) /*display #, mp, mdr.*/
end /*j*/ /* [↑] show MP and MDR for each #*/
say; target=5
say 'MDR first ' target " numbers that have a matching MDR"
say '═══ ═══════════════════════════════════════════════════'
do k=0 for 10; hits=0; _= /*show #'s that have an MDR of K.*/
do m=k until hits==target /*find target #s with an MDR of K*/
if word(mdr(m),2)\==k then iterate /*is the MDR what's wanted? */
hits=hits+1; _=space(_ m',') /*yes, we got a hit, add to list.*/
end /*m*/ /* [↑] built a list of MDRs = k */
say " "k': ['strip(_,,',')"]" /*display the K (mdr) and list.*/
end /*k*/ /* [↑] done with the K mdr list.*/
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────MDR subroutine──────────────────────*/
mdr: procedure; parse arg y; y=abs(y) /*get the number and find the MDR*/
do p=1 until y<10 /*find multiplicative digRoot (Y)*/
parse var y 1 r 2; do k=2 to length(y); r=r*substr(y,k,1); end; y=r
end /*p*/ /*wash, rinse, repeat ··· */
return p r /*return the persistence and MDR.*/

output   when using the default inputs:

 number   persistence   multiplicative digital root
────────  ───────────   ───────────────────────────
  123321       3                     8
    7739       3                     8
     893       3                     2
  899998       2                     0

MDR        first  5  numbers that have a matching MDR
═══   ═══════════════════════════════════════════════════
 0:     [0, 10, 20, 25, 30]
 1:     [1, 11, 111, 1111, 11111]
 2:     [2, 12, 21, 26, 34]
 3:     [3, 13, 31, 113, 131]
 4:     [4, 14, 22, 27, 39]
 5:     [5, 15, 35, 51, 53]
 6:     [6, 16, 23, 28, 32]
 7:     [7, 17, 71, 117, 171]
 8:     [8, 18, 24, 29, 36]
 9:     [9, 19, 33, 91, 119]

ultra-fast version[edit]

This fast version can handle a target of five hundred numbers with ease for the 2nd part of the task's requirement.

/*REXX pgm finds persistence and multiplicative digital root of some #'s*/
numeric digits 2000 /*increase the number of digits. */
parse arg target x; if \datatype(target,'W') then target=25 /*default?*/
if x='' then x=123321 7739 893 899998 /*use the defaults for X ? */
say center('number',8) ' persistence multiplicative digital root'
say copies('─' ,8) ' ─────────── ───────────────────────────'
/* [↑] title and separator. */
do j=1 for words(x); n=abs(word(x,j)) /*process each # in list.*/
parse value mdr(n) with mp mdr /*obtain the persistence and MDR.*/
say right(n,8) center(mp,13) center(mdr,30) /*display #, mp, mdr.*/
end /*j*/ /* [↑] show MP and MDR for each #*/
say /* [↓] show a blank & title line.*/
say 'MDR first ' target " numbers that have a matching MDR"
say '═══ ' copies("═",(target+(target+1)**2)%2) /*display a sep line.*/
 
do k=0 for 9; hits=0; _= /*show #'s that have an MDR of K.*/
if k==7 then [email protected]; else /*handle special seven case. */
 
do m=k until hits==target /*find target #s with an MDR of K*/
 ?=right(m,1) /*obtain right-most digit of M. */
if k\==0 then if ?==0 then iterate
if k==5 then if ?//2==0 then iterate
if k==1 then m=copies(1,hits+1)
else if mdr(m,1)\==k then iterate
hits=hits+1; _=space(_ m) /*yes, we got a hit, add to list.*/
 
if k==3 then do; o=strip(m,'T',1) /*strip trailing ones*/
if o==3 then m=copies(1,length(m))3 /*make new M. */
else do; t=pos(3,m)-1 /*position of 3*/
m=overlay(3,translate(m,1,3),t)
end /* [↑] shift the "3" 1 place left*/
m=m-1 /*adjust for DO index advancement*/
end /* [↑] a shortcut to do DO index*/
end /*m*/ /* [↑] built a list of MDRs = k */
 
say " "k': ['_"]" /*display the K (mdr) and list.*/
if k==3 then @7=translate(_,7,k) /*save for later, special 7 case.*/
end /*k*/ /* [↑] done with the K mdr list.*/
@.= /* [↓] handle MDR of 9 special. */
_=translate(@7,9,7) /*translate a string for MDR 9. */
@9=translate(_,,',') /*remove trailing commas from #'s*/
@3= /*assine null string before build*/
do j=1 for words(@9) /*process each number for MDR 9. */
_=space(translate(word(@9,j),,9),0) /*remove "9"s using SPACE(x,0)*/
L=length(_)+1 /*use a "fudged" length of the #.*/
new= /*this is the new numbers so far.*/
do k=0 for L; q=insert(3,_,k) /*insert the 1st "3" into the #*/
do i=k to L; z=insert(3,q,i) /* " " 2nd "3" " " "*/
if @.z\=='' then iterate /*if already define, ignore the #*/
@.z=z; new=z new /*define it, and then add to list*/
end /*i*/ /* [↑] end of 2nd insertion of 3*/
end /*k*/ /* [↑] " " 1st " " "*/
@3=space(@3 new) /*remove blanks, then add to list*/
end /*j*/ /* [↑] end of insertion of "3"s.*/
 
[email protected]; [email protected]; @= /*define three strings for merge.*/
/* [↓] merge two lists, 3s & 9s.*/
do while a1\=='' & a2\=='' /*process while the lists ¬empty.*/
x=word(a1,1); y=word(a2,1); if x=='' | y=='' then leave /*empty?*/
if x<y then do; @=@ x; a1=delword(a1,1,1); end /*add X.*/
else do; @=@ y; a2=delword(a2,1,1); end /*add Y.*/
end /*while ···*/ /* [+] only process just 'nuff. */
@=subword(@,1,target) /*elide the last trailing comma. */
say " "9': ['@"]" /*display the 9 (mdr) and list.*/
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────MDR subroutine──────────────────────*/
mdr: procedure; parse arg y,s /*get the number and find the MDR*/
do p=1 until y<10 /*find multiplicative digRoot (Y)*/
parse var y 1 r 2; do k=2 to length(y); r=r*substr(y,k,1); end; y=r
end /*p*/ /*wash, rinse, repeat ··· */
if s==1 then return r /*return multiplicative dig root.*/
return p r /*return the persistence and MDR.*/
output &nbsp' when the using the input of:   34
 number   persistence   multiplicative digital root
────────  ───────────   ───────────────────────────
  123321       3                     8
    7739       3                     8
     893       3                     2
  899998       2                     0

MDR       first  34  numbers that have a matching MDR
═══   ═════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════════
 0:     [0 10 20 25 30 40 45 50 52 54 55 56 58 59 60 65 69 70 78 80 85 87 90 95 96 100 101 102 103 104 105 106 107 108]
 1:     [1 11 111 1111 11111 111111 1111111 11111111 111111111 1111111111 11111111111 111111111111 1111111111111 11111111111111 111111111111111 1111111111111111 11111111111111111 111111111111111111 1111111111111111111 11111111111111111111 111111111111111111111 1111111111111111111111 11111111111111111111111 111111111111111111111111 1111111111111111111111111 11111111111111111111111111 111111111111111111111111111 1111111111111111111111111111 11111111111111111111111111111 111111111111111111111111111111 1111111111111111111111111111111 11111111111111111111111111111111 111111111111111111111111111111111 1111111111111111111111111111111111]
 2:     [2 12 21 26 34 37 43 62 73 112 121 126 134 137 143 162 173 211 216 223 232 261 278 279 287 297 299 314 317 322 341 367 369 371]
 3:     [3 13 31 113 131 311 1113 1131 1311 3111 11113 11131 11311 13111 31111 111113 111131 111311 113111 131111 311111 1111113 1111131 1111311 1113111 1131111 1311111 3111111 11111113 11111131 11111311 11113111 11131111 11311111]
 4:     [4 14 22 27 39 41 72 89 93 98 114 122 127 139 141 172 189 193 198 212 217 221 249 266 271 277 294 319 333 338 346 364 379 383]
 5:     [5 15 35 51 53 57 75 115 135 151 153 157 175 315 351 355 359 395 511 513 517 531 535 539 553 557 571 575 579 593 597 715 751 755]
 6:     [6 16 23 28 32 44 47 48 61 68 74 82 84 86 116 123 128 132 144 147 148 161 168 174 182 184 186 213 218 224 227 228 231 238]
 7:     [7 17 71 117 171 711 1117 1171 1711 7111 11117 11171 11711 17111 71111 111117 111171 111711 117111 171111 711111 1111117 1111171 1111711 1117111 1171111 1711111 7111111 11111117 11111171 11111711 11117111 11171111 11711111]
 8:     [8 18 24 29 36 38 42 46 49 63 64 66 67 76 77 79 81 83 88 92 94 97 99 118 124 129 136 138 142 146 149 163 164 166]
 9:     [9 19 33 91 119 133 191 313 331 911 1119 1133 1191 1313 1331 1911 3113 3131 3311 9111 11119 11133 11191 11313 11331 11911 13113 13131 13311 19111 31113 31131 31311 33111]

Ruby[edit]

Works with: Ruby version 2.1
def mdroot(n)
mdr, persist = n, 0
until mdr < 10 do
mdr = mdr.to_s.each_char.map(&:to_i).inject(:*)
persist += 1
end
[mdr, persist]
end
 
puts "Number: MDR MP", "====== === =="
[123321, 7739, 893, 899998].each{|n| puts "%6d:  %d  %2d" % [n, *mdroot(n)]}
 
counter = Hash.new{|h,k| h[k]=[]}
0.step do |i|
counter[mdroot(i).first] << i
break if counter.values.all?{|v| v.size >= 5 }
end
puts "", "MDR: [n0..n4]", "=== ========"
10.times{|i| puts "%3d: %p" % [i, counter[i].first(5)]}
Output:
Number: MDR  MP
======  ===  ==
123321:   8   3
  7739:   8   3
   893:   2   3
899998:   0   2

MDR: [n0..n4]
===  ========
  0: [0, 10, 20, 25, 30]
  1: [1, 11, 111, 1111, 11111]
  2: [2, 12, 21, 26, 34]
  3: [3, 13, 31, 113, 131]
  4: [4, 14, 22, 27, 39]
  5: [5, 15, 35, 51, 53]
  6: [6, 16, 23, 28, 32]
  7: [7, 17, 71, 117, 171]
  8: [8, 18, 24, 29, 36]
  9: [9, 19, 33, 91, 119]

Scala[edit]

Works with: Scala version 2.9.x
import Stream._
 
object MDR extends App {
 
def mdr(x: BigInt, base: Int = 10): (BigInt, Long) = {
def multiplyDigits(x: BigInt): BigInt = ((x.toString(base) map (_.asDigit)) :\ BigInt(1))(_*_)
def loop(p: BigInt, c: Long): (BigInt, Long) = if (p < base) (p, c) else loop(multiplyDigits(p), c+1)
loop(multiplyDigits(x), 1)
}
 
printf("%15s\t%10s\t%s\n","Number","MDR","MP")
printf("%15s\t%10s\t%s\n","======","===","==")
Seq[BigInt](123321, 7739, 893, 899998, BigInt("393900588225"), BigInt("999999999999")) foreach {x =>
val (s, c) = mdr(x)
printf("%15s\t%10s\t%2s\n",x,s,c)
}
println
 
val mdrs: Stream[Int] => Stream[(Int, BigInt)] = i => i map (x => (x, mdr(x)._1))
 
println("MDR: [n0..n4]")
println("==== ========")
((for {i <- 0 to 9} yield (mdrs(from(0)) take 11112 toList) filter {_._2 == i})
.map {_ take 5} map {xs => xs map {_._1}}).zipWithIndex
.foreach{p => printf("%3s: [%s]\n",p._2,p._1.mkString(", "))}
 
}
Output:
         Number        MDR      MP
         ======        ===      ==
         123321          8       3
           7739          8       3
            893          2       3
         899998          0       2
   393900588225          0       1
   999999999999          0       3

MDR: [n0..n4]
==== ========
  0: [0, 10, 20, 25, 30]
  1: [1, 11, 111, 1111, 11111]
  2: [2, 12, 21, 26, 34]
  3: [3, 13, 31, 113, 131]
  4: [4, 14, 22, 27, 39]
  5: [5, 15, 35, 51, 53]
  6: [6, 16, 23, 28, 32]
  7: [7, 17, 71, 117, 171]
  8: [8, 18, 24, 29, 36]
  9: [9, 19, 33, 91, 119]

Sidef[edit]

Translation of: Ruby
func mdroot(n) {
var (mdr, persist) = (n, 0)
while (mdr >= 10) {
mdr = mdr.digits.prod
++persist
}
[mdr, persist]
}
 
say "Number: MDR MP\n====== === =="
[123321, 7739, 893, 899998].each{|n| "%6d: %3d %3d\n" \
.printf(n, mdroot(n)...) }
 
var counter = Hash()
 
Inf.times { |j|
counter{mdroot(j).first} := [] << j
break if counter.values.all {|v| v.len >= 5 }
}
 
say "\nMDR: [n0..n4]\n=== ========"
10.times {|i| "%3d: %s\n".printf(i, counter{i}.first(5)) }
Output:
Number: MDR  MP
======  ===  ==
123321:   8   3
  7739:   8   3
   893:   2   3
899998:   0   2

MDR: [n0..n4]
===  ========
  0: [0, 10, 20, 25, 30]
  1: [1, 11, 111, 1111, 11111]
  2: [2, 12, 21, 26, 34]
  3: [3, 13, 31, 113, 131]
  4: [4, 14, 22, 27, 39]
  5: [5, 15, 35, 51, 53]
  6: [6, 16, 23, 28, 32]
  7: [7, 17, 71, 117, 171]
  8: [8, 18, 24, 29, 36]
  9: [9, 19, 33, 91, 119]

Tcl[edit]

proc mdr {n} {
if {$n < 0 || ![string is integer $n]} {
error "must be an integer"
}
for {set i 0} {$n > 9} {incr i} {
set n [tcl::mathop::* {*}[split $n ""]]
}
return [list $i $n]
}

Demonstrating:

puts "Number: MP MDR"
puts [regsub -all . "Number: MP MDR" -]
foreach n {123321 7739 893 899998} {
puts [format "%6d: %2d %3d" $n {*}[mdr $n]]
}
puts ""
# The longEnough variable counts how many roots have at least 5 values accumulated for them
for {set i [set longEnough 0]} {$longEnough < 10} {incr i} {
set root [lindex [mdr $i] 1]
if {[llength [lappend accum($root) $i]] == 5} {incr longEnough}
}
puts "MDR: \[n\u2080\u2026n\u2084\]"
puts [regsub -all . "MDR: \[n\u2080\u2026n\u2084\]" -]
for {set i 0} {$i < 10} {incr i} {
puts [format "%3d: (%s)" $i [join [lrange $accum($i) 0 4] ", "]]
}
Output:
Number: MP MDR
--------------
123321:  3   8
  7739:  3   8
   893:  3   2
899998:  2   0

MDR: [n₀…n₄]
------------
  0: (0, 10, 20, 25, 30)
  1: (1, 11, 111, 1111, 11111)
  2: (2, 12, 21, 26, 34)
  3: (3, 13, 31, 113, 131)
  4: (4, 14, 22, 27, 39)
  5: (5, 15, 35, 51, 53)
  6: (6, 16, 23, 28, 32)
  7: (7, 17, 71, 117, 171)
  8: (8, 18, 24, 29, 36)
  9: (9, 19, 33, 91, 119)

zkl[edit]

Translation of: Python
fcn mdroot(n){ // Multiplicative digital root
mdr := List(n);
while (mdr[-1] > 9){
mdr.append(mdr[-1].split().reduce('*,1));
}
return(mdr.len() - 1, mdr[-1]);
}
fcn mdroot(n){
count:=0; mdr:=n;
while(mdr > 9){
m:=mdr; digitsMul:=1;
while(m){
reg md;
m,md=m.divr(10);
digitsMul *= md;
}
mdr = digitsMul;
count += 1;
}
return(count, mdr);
}
println("Number:  (MP, MDR)\n=======  =========");
foreach n in (T(123321, 7739, 893, 899998))
{ println("%7,d: %s".fmt(n, mdroot(n))) }
 
table:=D([0..9].zip(fcn{List()}).walk()); // dictionary(0:List, 1:List, ...)
n  :=0;
while(table.values.filter(fcn(r){r.len()<5})){ // until each entry has >=5 values
mpersistence, mdr := mdroot(n);
table[mdr].append(n);
n += 1;
}
println("\nMP: [n0..n4]\n== ========");
foreach mp in (table.keys.sort()){
println("%2d: %s".fmt(mp, table[mp][0,5])); //print first five values
}
Output:
Number:  (MP, MDR)
=======  =========
123,321: L(3,8)
  7,739: L(3,8)
    893: L(3,2)
899,998: L(2,0)

MP: [n0..n4]
==  ========
 0: L(0,10,20,25,30)
 1: L(1,11,111,1111,11111)
 2: L(2,12,21,26,34)
 3: L(3,13,31,113,131)
 4: L(4,14,22,27,39)
 5: L(5,15,35,51,53)
 6: L(6,16,23,28,32)
 7: L(7,17,71,117,171)
 8: L(8,18,24,29,36)
 9: L(9,19,33,91,119)