# Largest number divisible by its digits

*is a*

**Largest number divisible by its digits****draft**programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

- Task

Find the largest base 10 integer whose digits are all different, and is evenly divisible by each of its individual digits.

These numbers are also known as **Lynch-Bell numbers**, numbers **n** such that the
(base ten) digits are all different (and do not include zero) and **n** is divisible by each of its individual digits.

For example: 135 is evenly divisible by 1, 3 and 5.

Note that the digit zero (0) can not be in the number as integer division by zero is undefined. The digits must all be unique so a base 10 number will have at most 9 digits.

Feel free to use analytics and clever algorithms to reduce the search space your example needs to visit, but it must do an actual search. (Don't just feed it the answer and verify it is correct.)

- Stretch goal

Do the same thing for hexadecimal.

- Also see

- The OEIS sequence: A115569: Lynch-Bell numbers.

## Contents

## C[edit]

### Base 10[edit]

The number can't contain 0 and 5, 0 is obvious, 5 because the number must end in 5 for it to be a multiple of that number and if that happens, all the even digits are ruled out which severely reduces the number's length since the other condition is that all digits must be unique. However, this means the number must be even and thus end only in 2,4,6,8. This speeds up the search by a factor of 2. The same approach when applied to hexadecimals takes a very long, long time.

/*Abhishek Ghosh, 8th November 2017*/

#include<stdio.h>

int main()

{

int num = 9876432,diff[] = {4,2,2,2},i,j,k=0;

char str[10];

start:snprintf(str,10,"%d",num);

for(i=0;str[i+1]!=00;i++){

if(str[i]=='0'||str[i]=='5'||num%(str[i]-'0')!=0){

num -= diff[k];

k = (k+1)%4;

goto start;

}

for(j=i+1;str[j]!=00;j++)

if(str[i]==str[j]){

num -= diff[k];

k = (k+1)%4;

goto start;

}

}

printf("Number found : %d",num);

return 0;

}

Output:

Number found : 9867312

## Haskell[edit]

### base 10[edit]

Using the analysis provided in the Perl 6 (base 10) example:

import Data.List (maximumBy, permutations, delete)

import Data.Ord (comparing)

unDigits :: [Int] -> Int

unDigits = foldl ((+) . (10 *)) 0

ds :: [Int]

ds = [1, 2, 3, 4, 6, 7, 8, 9] -- 0 (and thus 5) are both unworkable

lcmDigits :: Int

lcmDigits = foldr1 lcm ds -- 504

sevenDigits :: [[Int]]

sevenDigits = (`delete` ds) <$> [1, 4, 7] -- Dropping any one of these three

main :: IO ()

main =

print $

maximumBy

(comparing

(\x ->

if rem x lcmDigits == 0 -- Checking for divisibility by all digits

then x

else 0))

(unDigits <$> concat (permutations <$> sevenDigits))

- Output:

Test run from inside the Atom editor:

9867312 [Finished in 0.395s]

### base 16[edit]

First member of a descending sequence of multiples of 360360 that uses the full set of 15 digits when expressed in hex.

import Data.Set (fromList)

import Numeric (showHex)

lcmDigits :: Int

lcmDigits = foldr1 lcm [1 .. 15] -- 360360

upperLimit :: Int

upperLimit =

let allDigits = 0xfedcba987654321

in allDigits - rem allDigits lcmDigits

main :: IO ()

main =

print $

head

(filter ((15 ==) . length . fromList) $

(`showHex` []) <$> [upperLimit,upperLimit - lcmDigits .. 1])

Test run from inside the Atom editor:

"fedcb59726a1348" [Finished in 2.319s]

## Kotlin[edit]

Makes use of the Perl 6 entry's analysis:

### base 10[edit]

// version 1.1.4-3

fun Int.divByAll(digits: List<Char>) = digits.all { this % (it - '0') == 0 }

fun main(args: Array<String>) {

val magic = 9 * 8 * 7

val high = 9876432 / magic * magic

for (i in high downTo magic step magic) {

if (i % 10 == 0) continue // can't end in '0'

val s = i.toString()

if ('0' in s || '5' in s) continue // can't contain '0' or '5'

val sd = s.toCharArray().distinct()

if (sd.size != s.length) continue // digits must be unique

if (i.divByAll(sd)) {

println("Largest decimal number is $i")

return

}

}

}

- Output:

Largest decimal number is 9867312

### base 16[edit]

// version 1.1.4-3

fun Long.divByAll(digits: List<Char>) =

digits.all { this % (if (it <= '9') it - '0' else it - 'W') == 0L }

fun main(args: Array<String>) {

val magic = 15L * 14 * 13 * 12 * 11

val high = 0xfedcba987654321L / magic * magic

for (i in high downTo magic step magic) {

if (i % 16 == 0L) continue // can't end in '0'

val s = i.toString(16) // always generates lower case a-f

if ('0' in s) continue // can't contain '0'

val sd = s.toCharArray().distinct()

if (sd.size != s.length) continue // digits must be unique

if (i.divByAll(sd)) {

println("Largest hex number is ${i.toString(16)}")

return

}

}

}

- Output:

Largest hex number is fedcb59726a1348

## Perl 6[edit]

### Base 10[edit]

The number can not have a zero in it, that implies that it can not have a 5 either since if it has a 5, it must be divisible by 5, but the only numbers divisible by 5 end in 5 or 0. It can't be zero, and if it is odd, it can't be divisible by 2, 4, 6 or 8. So that leaves 98764321 as possible digits the number can contain. The sum of those 8 digits is not divisible by three so the largest possible integer must use no more than 7 of them (since 3, 6 and 9 would be eliminated). Strictly by removing possibilities that cannot possibly work we are down to at most 7 digits.

We can deduce that the digit that won't get used is one of 1, 4, or 7 since those are the only ones where the removal will yield a sum divisible by 3. It is *extremely* unlikely be 1, since EVERY number is divisible by 1. Removing it reduces the number of digits available but doesn't gain anything as far as divisibility. It is unlikely to be 7 since 7 is prime and can't be made up of multiples of other numbers. Practically though, the code to accommodate these observations is longer running and more complex than just brute-forcing it from here.

In order to accommodate the most possible digits, the number must be divisible by 7, 8 and 9. If that is true then it is automatically divisible by 2, 3, 4, & 6 as they can all be made from the combinations of multiples of 2 and 3 which are present in 8 & 9; so we'll only bother to check multiples of 9 * 8 * 7 or 504.

All these optimizations get the run time to well under 1 second.

my $magic-number = 9 * 8 * 7; # 504

my $div = 9876432 div $magic-number * $magic-number; # largest 7 digit multiple of 504 < 9876432

for $div, { $_ - $magic-number } ... * -> $test { # only generate multiples of 504

next if $test ~~ / <[05]> /; # skip numbers containing 0 or 5

next if $test ~~ / (.).*$0 /; # skip numbers with non unique digits

next unless all $test.comb.map: $test %% *; # skip numbers that don't divide evenly by all digits

say "Found $test"; # Found a solution, display it

for $test.comb {

printf "%s / %s = %s\n", $test, $_, $test / $_;

}

last

}

- Output:

Found 9867312 9867312 / 9 = 1096368 9867312 / 8 = 1233414 9867312 / 6 = 1644552 9867312 / 7 = 1409616 9867312 / 3 = 3289104 9867312 / 1 = 9867312 9867312 / 2 = 4933656

### Base 16[edit]

There are fewer analytical optimizations available for base 16. Other than 0, no digits can be ruled out so a much larger space must be searched. We'll start at the largest possible permutation (FEDCBA987654321) and work down so as soon as we find **a** solution, we know it is **the** solution.

my $hex = 'FEDCBA987654321'; # largest possible hex number

my $magic-number = [lcm] 1 .. 15; # find least common multiple

my $div = :16($hex) div $magic-number * $magic-number;

for $div, { $_ - $magic-number } ... 0 -> $num { # Only generate multiples of the lcm

my $test = $num.base(16);

next if $test ~~ / 0 /; # skip numbers containing 0

next if $test ~~ / (.).*$0 /; # skip numbers with non unique digits

say "Found $test"; # Found a solution, display it

say ' ' x 12, 'In base 16', ' ' x 36, 'In base 10';

for $test.comb {

printf "%s / %s = %s | %d / %2d = %19d\n",

$test, $_, ($num / :16($_)).base(16),

$num, :16($_), $num / :16($_);

}

last

}

- Output:

Found FEDCB59726A1348 In base 16 In base 10 FEDCB59726A1348 / F = 10FDA5B4BE4F038 | 1147797065081426760 / 15 = 76519804338761784 FEDCB59726A1348 / E = 1234561D150B83C | 1147797065081426760 / 14 = 81985504648673340 FEDCB59726A1348 / D = 139AD2E43E0C668 | 1147797065081426760 / 13 = 88292081929340520 FEDCB59726A1348 / C = 153D0F21EDE2C46 | 1147797065081426760 / 12 = 95649755423452230 FEDCB59726A1348 / B = 172B56538F25ED8 | 1147797065081426760 / 11 = 104345187734675160 FEDCB59726A1348 / 5 = 32F8F11E3AED0A8 | 1147797065081426760 / 5 = 229559413016285352 FEDCB59726A1348 / 9 = 1C5169829283B08 | 1147797065081426760 / 9 = 127533007231269640 FEDCB59726A1348 / 7 = 2468AC3A2A17078 | 1147797065081426760 / 7 = 163971009297346680 FEDCB59726A1348 / 2 = 7F6E5ACB93509A4 | 1147797065081426760 / 2 = 573898532540713380 FEDCB59726A1348 / 6 = 2A7A1E43DBC588C | 1147797065081426760 / 6 = 191299510846904460 FEDCB59726A1348 / A = 197C788F1D76854 | 1147797065081426760 / 10 = 114779706508142676 FEDCB59726A1348 / 1 = FEDCB59726A1348 | 1147797065081426760 / 1 = 1147797065081426760 FEDCB59726A1348 / 3 = 54F43C87B78B118 | 1147797065081426760 / 3 = 382599021693808920 FEDCB59726A1348 / 4 = 3FB72D65C9A84D2 | 1147797065081426760 / 4 = 286949266270356690 FEDCB59726A1348 / 8 = 1FDB96B2E4D4269 | 1147797065081426760 / 8 = 143474633135178345

## REXX[edit]

### base 10[edit]

This REXX version uses mostly the same logic and deductions that the **Perl 6** example does, but it performs

the tests in a different order for maximum speed.

The inner **do** loop is only executed a score of times; the 1^{st} **if** statement does the bulk of the eliminations.

/*REXX program finds the largest (decimal) integer divisible by all its decimal digits. */

$=7 * 8 * 9 /*a number that it must be divisible by*/

start=9876432 % $ * $ /*the number to start the sieving from.*/

t=0 /*the number of divisibility trials. */

do #=start by -$ /*search from largest number downwards.*/

if # // $ \==0 then iterate /*Not divisible? Then keep searching.*/

if verify(50,#,'M') \==0 then iterate /*does it contain a five or a zero? */

t=t+1 /*curiosity's sake, track # of trials. */

do j=1 for length(#) - 1 /*look for a possible duplicated digit.*/

if pos(substr(#,j,1),#,j+1)\==0 then iterate #

end /*j*/ /* [↑] Not unique? Then keep searching*/

/* [↓] superfluous, but check anyways.*/

do v=1 for length(#) /*verify the # is divisible by all digs*/

if # // substr(#,v,1) \==0 then iterate #

end /*v*/ /* [↑] ¬divisible? Then keep looking.*/

leave /*we found a number, so go display it. */

end /*#*/

say 'found ' # " (in " t ' trials)' /*stick a fork in it, we're all done. */

- output:

Timing note: execution time is under ** ^{1}/_{2}** millisecond (essentially not measurable in the granularity of the REXX timer under Microsoft Windows).

found 9867312 (in 11 trials)

### base 16[edit]

The "magic" number was expanded to handle hexadecimal numbers.

Note that **15*14*13*12*11** is the same as **13*11*9*8*7*5**.

/*REXX program finds the largest hexadecimal integer divisible by all its hex digits. */

numeric digits 20 /*be able to handle the large hex nums.*/

bigH= 'fedcba987654321' /*biggest hexadecimal number possible. */

bigN=x2d(bigH) /* " " " in decimal*/

$=15 * 14 * 13 * 12 * 11 /*a number that it must be divisible by*/

start=bigN % $ * $ /*the number to start the sieving from.*/

t=0 /*the number of divisibility trials. */

do #=start by -$ /*search from largest poss. # downwards*/

if # // $ \==0 then iterate /*Not divisible? Then keep searching.*/

h=d2x(#) /*convert decimal number to hexadecimal*/

if pos(0, h) \==0 then iterate /*does hexadecimal number contain a 0? */

t=t+1 /*curiosity's sake, track # of trials. */

do j=1 for length(h) - 1 /*look for a possible duplicated digit.*/

if pos(substr(h,j,1),h,j+1)\==0 then iterate #

end /*j*/ /* [↑] Not unique? Then keep searching*/

do v=1 for length(h) /*verify the # is divisible by all digs*/

if # // x2d(substr(h,v,1)) \==0 then iterate #

end /*v*/ /* [↑] ¬divisible? Then keep looking.*/

leave /*we found a number, so go display it. */

end /*#*/

say 'found ' h " (in " t ' trials)' /*stick a fork in it, we're all done. */

- output:

found FEDCB59726A1348 (in 287747 trials)

## Ring[edit]

# Project : Largest number divisible by its digits

# Date : 2017/09/22

# Author : Gal Zsolt (~ CalmoSoft ~)

# Email : <[email protected]ail.com>

for n = 9867000 to 9867400

numbers = list(9)

for t=1 to 9

numbers[t] = 0

next

flag = 1

flag2 = 1

flag3 = 1

str=string(n)

for m=1 to len(str)

if number(str[m]) > 0

numbers[number(str[m])] = numbers[number(str[m])] + 1

else

flag2 = 0

ok

next

if flag2 = 1

for p=1 to 9

if numbers[p] = 0 or numbers[p] = 1

else

flag = 0

ok

next

if flag = 1

for x=1 to len(str)

if n%(number(str[x])) != 0

flag3 = 0

ok

next

if flag3 = 1

see n + nl

ok

ok

ok

next

Output:

9867312

## zkl[edit]

### base 10[edit]

const magic_number=9*8*7; # 504

const div=9876432 / magic_number * magic_number; #largest 7 digit multiple of 504 < 9876432

foreach test in ([div..0,-magic_number]){

text:=test.toString();

if(text.holds("0","5")) continue; # skip numbers containing 0 or 5

if(text.unique().len()!=text.len()) continue; # skip numbers with non unique digits

if(test.split().filter1('%.fp(test))) continue; # skip numbers that don't divide evenly by all digits

println("Found ",test); # Found a solution, display it

foreach d in (test.split()){

println("%s / %s = %s".fmt(test,d, test/d));

}

break;

}

- Output:

Found 9867312 9867312 / 9 = 1096368 9867312 / 8 = 1233414 9867312 / 6 = 1644552 9867312 / 7 = 1409616 9867312 / 3 = 3289104 9867312 / 1 = 9867312 9867312 / 2 = 4933656

### base 16[edit]

const bigN=0xfedcba987654321; // biggest hexadecimal number possible.

lcm:=lcmNs([1..15]); // 360360, smallest # that will divide answer

upperLimit:=bigN - bigN%lcm; // start at a mulitple of whatever the answer is

foreach test in ([upperLimit..1,-lcm]){

text:=test.toString(16);

if(15!=text.unique().len()) continue;

println(text);

break;

}

fcn lcmNs(ns){ ns.reduce(fcn(m,n){ (m*n).abs()/m.gcd(n) }) }

- Output:

fedcb59726a1348