Is a number divisible by each of its digits?

Question

My friend and I were working on a lab in our AP Computer Science class and decided to code golf one one the problems since we still had half the class free after we finished. Here is the question:

Given a number n, is n divisible by each of its digits?

For example, 128 will pass this test- it is divisible by 1,2, and 8. Any numbers with a zero automatically disqualify the number. While you may use other languages and post solutions with them if you like, we are most interested in seeing how compact people can make the program in Java, as that is the language we use in the class. So far, we both have 51. Here is my current code:

public boolean dividesSelf(int n){for(int p=n;n%10>0;)n/=p%(n%10)>0?.1:10;return n<1;}
// 51 characters

// Breakdown:
// for(int p=n;         Saves one semicolon to put declaration into for loop
// n%10>0;)             Basic check-for-zero
// n/=                  Pretty simple, discarding one number off of n at a time
// p%(n%10)>0?          If p (the given value) is not divisible by n%10 (the current digit)...
// .1:10;               Divide by .1 (multiply by 10) so it fails the check next iteration. If it is divisible, divide by 10 to truncate the last digit
// return n<1           If the number was fully divisible, every digit would be truncated, and n would be 0. Else, there would still be non-zero digits.

Requirements

The method signature can be whatever you want. Just count the function body. Make sure, though, that the method returns a boolean value and only passes in one numeric parameter (not a string).

The code must be able to pass all of these cases (in order to stay true to the directions of the original question, only boolean true and false values count if the language supports booleans. If and only if your language does not have boolean variables you may represent false with 0 and true with any nonzero integer (preferably 1 or -1):

128 -> true
 12 -> true
120 -> false
122 -> true
 13 -> false
 32 -> false
 22 -> true
 42 -> false
212 -> true
213 -> false
162 -> true
204 -> false

Also, we didn't count whitespace, so feel free to do the same, unless the whitespace is essential to the working of the program (so newlines in Java don't count, but a single space between int and x=1 does count.) Good luck!

Answer 1

Perl 6, 13

sub golf($_) {
   $_%%.comb.all
}

Uses the implicit variable $_ — $_ %% .comb.all is equivalent to $_ %% all($_.comb). %% is the "is divisible" operator, and comb with no additional argument returns a list of the characters in a string. As an example, if the argument is 123, then the function evaluates

123 %% all(123.comb)

which is

123 %% all(1, 2, 3)

junction autothreading makes it

all(123 %% 1, 123 %% 2, 123 %% 3)

which is

all(True, False, True)

which is false in boolean context because it's an "all" junction and clearly not all of its elements are true.

It should be possible to coerce the return value to Bool and hide the junction-ness from callers by making the function signature sub golf($_ --> Bool()), but coercions in function signatures don't work yet in Rakudo. The return value is still correctly true or false, it's just not True or False.

Answer 2

C# and System.Linq - 26 / 40

Per the rules, not counting the method declaration itself.

bool dividesSelf(int i) { 
    return(i+"").All(d=>i%(d-48d)<1);
}

Showing that once again, C# is the superior choice when Java is under consideration... I kid, I kid!

Unfortunately, this function (and many in other answers) will not produce correct results for negative input. We can fix this, but the solution loses a lot of its charm (and grows to 46 characters in length):

return(i+"").All(d=>d>48&&i%(d-48)==0||d==45);

Edit: shaved off one character with Tim's suggestion.

Edit: with the introduction of expression-bodied members in C# 6, we can pare this down further by cutting out the return:

bool dividesSelf(int i) =>
    (i+"").All(d=>i%(d-48d)<1);

for a total of 26 characters (in my opinion, the => should not be included any more than braces would be). The version handling negative numbers can be similarly shortened.

Answer 3

APL (13 11)

(apparently the brackets don't count)

{0∧.=⍵|⍨⍎¨⍕⍵}

Explanation:

• ⍎¨⍕⍵: evaluate each character in the string representation of ⍵
• ⍵|⍨: for each of those, find the modulo of it and ⍵
• 0∧.=: see whether all of those are equal to 0

Testcases:

      N,[.5] {0∧.=⍵|⍨⍎¨⍕⍵} ¨ N←128 12 120 122 13 32 22 42 212 213 162 204
128 12 120 122 13 32 22 42 212 213 162 204
  1  1   0   1  0  0  1  0   1   0   1   0

Answer 4

Perl - 27 bytes

sub dividesSelf{
    $_=pop;s/./!$&||$_%$&/ger<1
}

Not counting the function signature, as instructed.

Sample Usage:

use Data::Dump qw(dump);
for $i (128, 12, 120, 122, 13, 32, 22, 42, 212, 213, 162, 204) {
  printf "%3d -> %s\n", $i, dump(dividesSelf $i);
}

Sample Output:

128 -> 1
 12 -> 1
120 -> ""
122 -> 1
 13 -> ""
 32 -> ""
 22 -> 1
 42 -> ""
212 -> 1
213 -> ""
162 -> 1
204 -> ""

---

Addressing the problem specification: "Only boolean true and false values count. Truthy/falsey values do not count."

use Data::Dump qw(dump);
dump(1 == 1);
dump(0 == 1);

Outputs:

1
""

'True' and 'False' are defined as 1 and "".

Erratum:
As Brad Gilbert rightly points out, perl defines true as a scalar which is both the integer 1 and the string "1" simultaneously, and false as a scalar which is both the integer 0 and the string "" simultaneously.

Answer 5

Python 2: 43 chars

f=lambda n:any(n%(int(d)or.3)for d in`n`)<1

Checks whether the number has any nonzero remainders modulo its digits, and outputs the negation of that. Zero digits are handled strangely: since computing %0 causes an error, digits of 0 are replaced with .3, which seems to always give a nonzero result due to floating point inaccuracies.

The function body is 32 chars.

Answer 6

CJam, 11 10 bytes

{
    _Ab:df%:+!
}:F;

This defines a function named F and discards the block from the stack.

Try it online.

Test cases

$ cjam <(echo '{_Ab:df%:+!}:F;[128 12 120 122 13 32 22 42 212 213 162 204]{F}%p')
[1 1 0 1 0 0 1 0 1 0 1 0]

How it works

_      " Copy the integer on the stack.                                          ";
Ab     " Push the array of its digits in base 10.                                ";
:d     " Cast each digit to Double.                                              ";
f%     " Take the integer on the stack modulus each of its digits.               ";
:+     " Add the results.                                                        ";
!      " Push the logical NOT of the sum.                                        ";

Answer 7

JavaScript ES6, 39 32 28 bytes

v=>[...""+v].every(x=>v%x<1)

Thanks core1024 for the suggestion to replace (""+v).split("") with [...""+v], and openorclose for suggesting the use of every function.

The answer currently doesn't contain one bit of my code :O

Previous solution

v=>[...""+v].filter(x=>v%x|!+x)==""

=="" is not a valid way to check if an array is empty, since [""]=="" returns true, but the array is guarantee to contain non-empty string, so it works here.

The rest are quite standard shorthand type conversion in JavaScript.

Answer 8

Java 8, 46 Bytes (method body)

Using Jeroen Mostert's converting to double trick.

public static boolean dividesSelf(int n) {
    return(""+n).chars().allMatch(x->n%(x-48d)<1);
}

Answer 9

Pyth, 12 bytes

!f|!vT%vzvTz

This filters the characters in the string for being either zero (!vT) or not dividing the input (%vzvT), then takes the logical not of the resulting list.

Try it here.

Answer 10

Ruby, 44 bytes (function body: 37)

Probably has potential to be golfed further.

f=->n{n.to_s.chars.all?{|x|x>?0&&n%x.hex<1}}

Input taken through function f. Example usage:

f[128] # => true
f[12]  # => true
f[120] # => false
...

Answer 11

Python - 59 50 49 47 bytes

f=lambda n:all(c>'0'and 0==n%int(c)for c in`n`)

I'm sure there's a faster way... oh well.

Edit - Thanks to FryAmTheEggman for the golfing tips.

Edit 2 - FryAmTheEggman may as well have written this at this point, oops

Edit 3 - Hands up if you didn't even know genexps were a thing. ...Just me?

Answer 12

Pyth 11

!f%Q|vT.3`Q

This combines @isaacg's and @xnor's answers. It filters out digits from the input by checking the value of input % (eval(current_digit) or .3). Then it checks if the resulting string is empty or not.

Came across another couple same-length variants:

!f%Q|T.3jQT
!f|!T%QTjQT

Try it online.

Answer 13

Bash + coreutils, 44 bytes

The full function definition is:

f()((`tr 0-9 \10<<<$1``sed "s/./||$1%&/g"<<<$1`))

I'm not sure how to score this as normally shell functions use a single set of {} or () to contain the function body. I found here I could also use double (()) to contain the function body which causes an arithmetic expansion which is what I need here. So for now I am counting just one pair of those brackets - further discussion of this is welcome.

Output:

$ for i in 128 12 120 122 13 32 22 42 212 213 162 204; do f $i; printf "%d " $?; done
1 1 0 1 0 0 1 0 1 0 1 0 $
$

Answer 14

J - 14 char

The function body is the portion after the =:. If we want to minimize the character count for the whole function, that's the 15 char */@(0=,.&.":|]).

f=:0*/@:=,.&.":|]

,.&.": is the shortest way in J to expand as number into a list of its decimal digits: convert to string, separate the digits, and convert each digit back into a number. ,.&.":|] takes the input number (]) modulo (|) those digits. 0*/@:= returns true if all the results were 0, else gives a false.

   f 162
1
   f every 204 212 213
0 1 0

Answer 15

Java - 121 102 97 79 78 bytes

I just know this will get clobbered later. Oh well.

boolean b(int a){int m=10,j,t=1;for(;m<a*10;m*=10){j=10*(a%m)/m;if(j==0||a%j>0)t=0;}return t>0;}

I'll be back.

Answer 16

Haskell - 100 54 38

f x=all(\y->y>'0'&&x`mod`read[y]<1)$show x

Still learning, critiques appreciated

Answer 17

CJam, 15 bytes

{_Abf{_{%}1?}1b!}

{} is the closest thing to a function in CJam. I am just counting the body of the function

Use it like this:

128{_Abf{_{%}1?}1b!}~

To get either 1 (if the number is divisible) or 0 (if the number is not divisible by its digits).

Try it online here

Explanation

_Ab                "Copy the number and split it to its digits";
   f{      }       "For each digit, run this code block on the number";
     _{%}1?        "If the digit is 0, put 1, otherwise perform number modulus digit";
            1b     "We now have an array of modulus corresponding to each digit. Sum it up";
              !    "Negate the sum. If all digits were divisible, sum of modules will be"
                   "0, thus answer should be 1 and vice versa";

Answer 18

C89, 43 bytes

unsigned char d(int n, int c) {
        int a=n%10;return!n||a&&!(c%a)&&d(n/10,c);
}

C89 doesn't have a boolean type. Hope that works. Also I used a second parameter to pass a copy of the original number through the stack, but the definition can be anything. To get the correct result you just have to call the function with the same value for both parameters (d(128, 128)).

EDIT: Applied suggested edits by an anonymous user

Answer 19

C / C++, 58 bytes (44 in body)

Invokes Undefined Behaviour (see comments)

int d(int i){int j=i;while(i&&!(j%(i%10)))i/=10;return!i;}

true and false are 1 and 0, but feel free to add one character to the signature to return a bool.

And for fun, a recursive version which is smaller if you allow calls of the form r(128,128)

Edit: Now disallowed by the rules:

C / C++, 53 bytes (33 in body)

int r(int i,int j){return!i||!(j%(i%10))&&r(i/10,j);}

Answer 20

CJam, 15 bytes

{_Abf{_g{%}*}:|!}

This is a block, the closest thing to a function in CJam. I'm only counting the body (i.e. omitting the braces). You can use it as follows:

128{_Abf{_g{%}*}:|!}~

Or if you want to test a series of inputs, you can do

[128 12 120 122 13 32 22 42 212 213 162 204]{{_Abf{_g{%}*}:|!}~}%

The block leaves 0 (falsy) or 1 (truthy) on the stack to indicate the result. (CJam doesn't have a Boolean type.)

Test it here.

Explanation:

_               "Duplicate input.";
 Ab             "Get base-10 digits.";
   f{      }    "This maps the block onto the list of digits, supplying the input each time.";
     _g         "Duplicate digit, get signum S (0 or 1).";
       { }*     "Repeat this block S times.";
        %       "Take input modulo digit.";
                "This leaves an array of zeroes for divisible digits, non-zeroes
                 for non-divisible digits, and non-zero junk for zeroes.";
            :|  "Fold OR onto this list. One could also sum the list with :+";
              ! "Logical NOT. Turns 0 into 1, and non-zero values into 0.";

Alternative, also 15 bytes

{:XAb{X\_X)?%},!}

Explanation

:X              "Store input in X.";
  Ab            "Get base-10 digits.";
    {       },  "Filter this list by the result of the block.";
     X\         "Push another copy of X, swap with digit.";
       _        "Duplicate digit.";
        X)      "Push X+1.";
          ?     "Select digit itself or X+1, depending on whether digit is 0 or not.";
           %    "Take modulo. X%(X+1) will always be nonzero for positive integers.";
              ! "Logical NOT. Turns an empty list into 1 and a non-empty list into 0.";

Answer 21

C11 - 44 Bytes in function body

Another C version, non recursive and without a floating point exception.

bool digit_multiple(int i)
{
    for(int n=i;i%10&&n%(i%10)<1;i/=10);return!i;
}

This will also work in C++, Java, and most other C-like languages.

Edited to include the improvement of primo's comment.

Answer 22

Julia 32 25 23

Improved using digits

Also fixes problem with negative numbers

selfDivides(x)=sum(x%digits(x).^1.)==0

Old method

All digits divide if the sum of all the remainders is 0. Like others, has a problem with negative numbers.

selfDivides(x)=sum(x.%(Float64["$x"...]-48))==0

Output

[selfDivides(x) for x in [128,12,120,122,13,32,22,42,212,213,162,204]]
12-element Array{Any,1}:
  true
  true
 false
  true
 false
 false
  true
 false
  true
 false
  true
 false

Improved method also handles BigInt

selfDivides(BigInt(11111111111111111111111111111111111111112))
true

however

selfDivides(BigInt(11111111111111111111111111111111111111113))
false

because

BigInt(11111111111111111111111111111111111111113) %3
1

Answer 23

R: 72 67 65

The function

f<-function(a)!(anyNA(a%%(d=as.double(strsplit(paste0(a),"")[[1]])))|sum(a%%d))

Thanks to @AlexA and @plannapus for the savings

Test run

i=c(128,12,120,122,13,32,22,42,212,213,162,204)
for(a in i){print(f(a))}
[1] TRUE
[1] TRUE
[1] FALSE
[1] TRUE
[1] FALSE
[1] FALSE
[1] TRUE
[1] FALSE
[1] TRUE
[1] FALSE
[1] TRUE
[1] FALSE

Answer 24

05AB1E, 6 5 bytes

ѨK0‹

Try it online!

Explanation

ѨK0‹  full program with implicit input  [213]
Ñ      push divisors                     [1, 3, 71, 213]
 ¨     remove the last element           [1, 3, 71]
  K    push a without b                  [2]
   0   push 0                            [2] [0]
    ‹  push a < b                        [0]
       output implicitly

K uses the implicit input [213] as a and the current stack [1, 3, 71] as b.

Answer 25

Stacked, 11 bytes in the function body

[:10 tb\|all]

Try it online!

Takes a number from the stack and returns 1 if they are all divisible or 0 otherwise.

Explanation

[:10 tb\|all]
[           ]   anonymous function (the function body)
 :10 tb         push digits of the number (*t*o *b*ase 10)
       \|       vectorized "divides" test
         all    ensure all are truthy (i.e. not 0)

Answer 26

GNU Awk: 53 characters

The counted part:

for(;++i<=split($1,a,//);)r=r||!a[i]||v%a[i];return!r

The entire function:

function self_divisible(v, i, r)
{
    for (; ++i <= split($1, a, //); )
        r = r || ! a[i] || v % a[i]

    return ! r
}

As Awk has no boolean values, returns 1 tor true and 0 for false.

Answer 27

JavaScript (ES6) 30

Function with one numeric parameter. Using % and subtraction, no need to special case '0' because 0%0 is NaN in JavaScript.

Edit Saved 1 char thx DocMax

F=n=>[for(d of t=n+'')t-=n%d]&&t==n 

Just for fun, abusing the rule about not counting function signature, 4

Check=(n,t=n+'',q=[for(d of t)n-=t%d])=>t==n

Test In FireFox/FireBug console

console.log([128, 12, 120, 122, 13, 32, 22, 42, 212, 213, 162, 204]
.map(x=>+x + ' -> ' + F(x)).join('\n'))

Output

128 -> true
12 -> true
120 -> false
122 -> true
13 -> false
32 -> false
22 -> true
42 -> false
212 -> true
213 -> false
162 -> true
204 -> false

Answer 28

PHP: 85 bytes (64 bytes on the body)

For this function to work, simply pass a string or a number.

0 will correctly return false.

The code:

function f($n,$i=0){for($n.='';$n[$i]&&$t=!($n%$n[$i++]););return$t&&$i==strlen($n);}

Please, DO NOT SET THE 2ND PARAMETER!

Javascript: 76 bytes (61 bytes on the body)

This is a rewrite of the previous function.

Not much changed between both versions.

Here is the code:

function f(n){for(i=0,n+='';n[i]/1&&(t=!(n%n[i++])););return t&&i==n.length}

---

Polyglot: Javascript+PHP 187 217 bytes (76 84 bytes without boilerplate):

Why I made it?

Because of reason and maybe because I can!

Just ignore the error on PHP: it works anyway!
No longer needed, this was fixed by removing 3 bytes.

Here is the masterpiece:

if('\0'=="\0"){function strlen($s){return $s['length'];}}
function toString(){return'';}
function f($n){for($i=0,$n=$n.toString();$n[$i]/1&&($t=!($n%$n[$i++])););return $t&&$i==strlen($n);}

You can run this code both on your console and on a PHP interpreter!

---

Old version:

if('\0'=="\0"){function strlen($s){return $s['length'];}}
function s($s){return('\0'=="\0")?$s+'':str_replace('','',$s);}
function f($n,$i){for($i=0,$n=s($n);$n[$i]/1&&($t=!($n%$n[$i++])););return $t&&$i==strlen($n);}

Answer 29

Octave, 33 (39 including function setup)

Using numeric-to-matrix conversion:

f=@(a)sum(mod(a./(num2str(a)-48),1))==0

Divide number elementwise by matrix X, where X is made by converting number to string and subtracting 48 to go from ASCII values to numbers again. Take modulo 1 to get decimal part of each division, confirm that all of these are zero (if any are NaN because of /0, the sum will be NaN and hence not zero).

Sample input using www.octave-online.net:

f=@(a)sum(mod(a./(num2str(a)-48),1))==0
for j=[128,12,120,122,13,32,22,42,212,213,162,204]
f(j)
end

Output:

ans =  1
ans =  1
ans = 0
ans =  1
ans = 0
ans = 0
ans =  1
ans = 0
ans =  1
ans = 0
ans =  1
ans = 0

Answer 30

PHP - 74 71 64 Characters

Golfed:

function t($n){while($n>1){if(!($b=$n%10)||($n%$b)){return 0;}$n/=10;}return 1;}

Less Golfed:

function t($n){
    while($n>1){
        if( !($b=$n%10) || ($n%$b) )
            { return 0; }
        $n/=10;
    }
    return 1;
}

Test Results:

(Code)

$ans = array(128,12,120,122,13,32,22,42,212,213,162,204);
foreach($ans as $a)
{ echo "$a -> ".(t($a)?"True":"False").PHP_EOL; }

(Output)

128 -> True
12 -> True
120 -> False
122 -> True
13 -> False
32 -> True
22 -> True
42 -> True
212 -> True
213 -> True
162 -> False
204 -> False

Answer 31

MATLAB - 39 characters

function [b] = dividesSelf(i)
b=all(~mod(i,sscanf(num2str(i),'%1d')))
end

Answer 32

BASH - 117 characters

f(){ [[ $1 =~ 0 ]]&& return 0 || r=;n=$1;for((i=0;i<${#n};i++));do r=$(($r+${n}%${n:$i:1}));done;return $(($r==0));}

tests

for N in 128 12 120 122 13 32 22 42 212 213 162 204; do
  f $N
  echo "${N} ->  $?"
done

128 ->  1
12 ->  1
120 ->  0
122 ->  1
13 ->  0
32 ->  0
22 ->  1
42 ->  0
212 ->  1
213 ->  0
162 ->  1
204 ->  0

Answer 33

Javascript ES6, 33

n=>(""+n).replace(/./g,x=>n%x)==0

Test:

f=n=>(""+n).replace(/./g,x=>n%x)==0
console.log(
           [128, 12,  120,  122, 13,   32,   22,   42,  212, 213,  162, 204]
.map(f) == "true,true,false,true,false,false,true,false,true,false,true,false"
)

Answer 34

PHP: 45 Characters

The character count is for the body of the function.

It is necessary to only pass the first parameter.

function t($n, $k=true, $s='str_split'){foreach($s($n)as$b)$k=$k&&$n%$b===0;return$k;}

Answer 35

K (oK), 13 bytes

Solution:

f:{~+/(48!$x)!'x}

13 bytes excludes function definition (+4) per specification.

Try it online!

Examples:

f 1024
0
f 128
1
f 333
1
f 334
0

Explanation:

Evaluation is performed right-to-left.

Convert (e.g.) 128 => 1 2 8. 128 mod each 1 2 8 => 0 0 0. Sum of this is 0. Not of this is 1:

f:{~+/(48!$x)!'x} / the solution including function definition
f:{             } / assign lambda function to f
               x  / implicit single parameter
             !'   / mod (!) each-left with each-right
      (     )     / do this together
          $x      / string x, 128 => "128"
       48!        / mod ascii values with 48, "128" => 1 2 8
    +/            / sum over results
   ~              / not, 0 => 1, anything else => 0

Answer 36

PHP, 44 bytes

function f($n,$s=str_split){
    foreach($s($n)as$d)$f|=!$d||$n%$d;return!$f;
}

Try it online.

Answer 37

Jelly, 3 bytes

ọDẠ

Try it online!

This is a link, i.e. a function in Jelly. Since Jelly doesn't have specific boolean values, this returns 1 for True and 0 for False.

Answer 38

Python 2: 36 38 Bytes

Excluding f=lambda n:

f=lambda n:all([not n%(int(a)or n-1)for a in`n`])

Similar to the other python solution. Works because x or 0 returns x if x != 0.

Maybe some edge cases around 0-3 or something.

Edit: Didn't work, fixed it

Answer 39

J, 18 Bytes (21 with function definition)

d=:0:=+/@:(|~"."0@":)

Explanation:

d=:                     | Define d
                  ":    | Format into string
             "."0       | Turn into array of digits
           |~           | Starting Number mod each element
      +/                | Sum
   0:=                  | Test if it’s equal 0. Returns 1 for true, 0 for false

Other characters are for ensuring functions compose properly.

Answer 40

Python 2, ~85 ~83 characters (~75 ~73 just the function body)

Staying all numeric; count includes newlines and spacing.

def t(n):
 r=n
 while r:
  r,i=divmod(r,10)
  if not i or n%i:return 0>1
 return 1>0

assert(t(128)==True)
assert(t(12)==True)
assert(t(120)==False)
assert(t(122)==True)
assert(t(13)==False)
assert(t(32)==False)
assert(t(22)==True)
assert(t(42)==False)
assert(t(212)==True)
assert(t(213)==False)
assert(t(162)==True)
assert(t(204)==False)

Alternative, 88 characters in the function body:

def t(n):return '0' not in `n` and not reduce(lambda a,b:a or b,[divmod(n,int(x))[1] for x in `n`])

The approach feels shorter with map/reduce but with Python code it isn't.

Answer 41

J (16)

+/@:|~_1&(10&#.)

Returns 0 when true, another number otherwise

Answer 42

PHP - 91 characters

function divides($n) {
    foreach(array_map('intval', str_split($n))as$i)if($i==0||$n%$i!=0)return false;return true;
}

foreach ([128, 12, 120, 122, 13, 32, 22, 42, 212, 213, 162, 204] as $n) {
    echo "\n" . $n . ' -> ' . divides($n);
}

Answer 43

Mathematica, 53 bytes (46 byte body)

f[x_]:=AllTrue[IntegerDigits@x,#!=0&&Divisible[x,#]&]

Answer 44

PHP, 59 Characters

Where $n is the number you're checking

foreach(str_split($n)as$v)if($n%$v!=0){echo 0;exit;}echo 1;

Answer 45

Haskell, 55

let f x =(\s->all(>'0')s||all((<1).(x`mod`).read.(:[]))s)$show x

Counting characters after the = sign as per the rules.

Answer 46

Python 2, 41 bytes

Size not including f=lambda j:.

f=lambda j:all(j%int(i)==0if i>"0"else 0for i in`j`)

Argument supplied must be an integer.

Basically what this does is iterates through the integer (after converting it to a string) and checks the divisibility.

Answer 47

Java 104

Uses recursion. This test (i==0) is required to stop divide by zero error.

public boolean d(int n,int k)
{
int i=k%10;return (k<10)?(k>0)&&(n%k==0):(i==0)?false:(n%i==0)&d(n,k/10);  
}

Answer 48

Batch - 202 Bytes

Takes input from stdin - C:\div.bat 128.

@echo off&setLocal EnableDelayedExpansion&set s=%1&set c=%1&set l=0
:c
if defined s set/Al+=1&set "s=%s:~1%"&goto c
for /l %%a in (1,1,!l!)do set/aa=%c%%%%%a&if !a!==1 echo false&goto :EOF
echo true

Answer 49

Java - (64)

This is the shortest I can come up with:

boolean t(int a) {
    int x=a,t=0;do t+=x%10>0?a%(x%10):0;while((x/=10)>0);return t<1;
}

If you don't count the method overhead, just the body, it scores 64.

Answer 50

Scala, 38 characters

def isDivisibleByAllDigits(n: Int) = n.toString.forall(d=>d>48&&n%(d-48)<1)

Answer 51

ES6, not exactly a golf, just screwing around with the Y-combinator

let dividesSelf = v =>
  !(l =>
    (f => f(f))
    (f => l(n => f(f)(n))))
  (f => n => n + v % n)
  (v >> 1);

A little minified, 52 characters long

let dividesSelf = v =>
  !(l=>(f=>f(f))(f=>l(n=>f(f)(n))))(f=>n=>n+v%n)(v>>1)

The fact I beat about half of the answers is surprising.

Answer 52

Go - 47 chars

func dividesSelf(i int) bool {
    for n:=i;i%10>0&&n%(i%10)<1;i/=10{};return i==0
}

Runnable on Go Playground: http://play.golang.org/p/67DQXWnwOT

Answer 53

Bash, 49

Exits with 0 if the number is divisible, other exit code (and a warning) if not.

f () { [ `<<<$1 sed "s/./+$1%\0/g"|cut -c2-|bc` -eq 0 ]; }

Answer 54

Clojure, 71 bytes

(fn[n](every? integer?(map #(if(=\0 %1)1.1(/ n(- 48(int %1))))(str n))))

Answer 55

JAGL Alpha 1.2 - 24 bytes

Obviously this wouldn't count in a real contest, because the language came out after the question, but I'm still gonna answer just for fun. Assumes that the number is the only value on the stack, and sets the top value of the stack to 1 or 0, depending on whether the number is divisible by its digits or not.

dg{i}/d0en{{SdcS%n}/A}Sf

Explanation:

dg                         Duplicate number and convert to string
  {i}/                     Convert that to an array of digits
      0en                  If 0 is in the array, push 0, else push 1
         {                 Block to be run if 0 not in array
          {SdcS            Swap, duplicate, cycle, swap
               %n}         Push 1 if divisible by digit, 0 otherwise
                  /A       Map over list and push 1 if all values are 1, otherwise 0
                    }Sf    Run block if 0 not in array

Answer 56

JavaScript, 89 81 70 bytes

function d(n){a=(n+"").split("");for(i in a)if(n%a[i]||1/0==n/a[i])return!1;return!0}

Answer 57

Javascript, 82 81 80 bytes

a=prompt(q=1);b=a.split('');while(d=b.pop())q=q&!(a%d);alert(a.indexOf(0)+1?0:q)

Answer 58

R, 40 chars

f=function(n)all(n%%scan(t=gsub("(.)","\\1 ",n))%in%0)

%% is vectorized. scan(t=gsub("(.)","\\1 ",n) is a trick that I stole from an old answer by flodel, to split a number into its digits. %in%0 instead of simply negating the results allows for detection of NA (resulting from mod 0; NA==0 gives NA, !NA gives NA but NA%in%0 gives FALSE).

> f(120)
Read 3 items
[1] FALSE
> f(128)
Read 3 items
[1] TRUE
> f(12)
Read 2 items
[1] TRUE
> f(213)
Read 3 items
[1] FALSE
> f(212)
Read 3 items
[1] TRUE
> f(42)
Read 2 items
[1] FALSE

Answer 59

TI-BASIC, 29

Adapted from my answer to Generate Monday Numbers.

Input X
int(10fPart(X10^(-randIntNoRep(1,1+int(log(X
not(max(remainder(X,Ans+2Xnot(Ans

To disqualify numbers that contain zeroes, I replace all zeroes with 2X, which will not divide X.

I don't count the Input X in the code size, because it is analogous to a method signature.

Answer 60

Hassium, 130 Bytes

func i(n){m=n.toString();if(m.contains("0"))return false;foreach(d in m)if(n%d.toString().toDouble()!=0)return false;return true;}

Run and see expanded version here

Answer 61

Java 7, Score = 96 76

Since I can't compete with your number based method, I used a character based method instead. a has the input integer in it.

int c=0;for(char e:(""+a).toCharArray())c+=(e-=-48)<1||a%e>0?1:0;return c<1;

Here it is golfed indented with structure code:

public class T{

    public static void main(String[] args){
        System.out.println(new T().test(Integer.parseInt(args[0])));
    }

    boolean test(int a){
        int c=0;
        for(char e:(""+a).toCharArray())
            c+=(e-=-48)<1||a%e>0?1:0;
        return c<1;
    }

}

Here it is completely expanded:

public class Test{

    public static void main(String[] args){
        System.out.println(new Test().test(Integer.parseInt(args[0])));
    }

    boolean test(int number){
        String numberString = "" + number;
        int failed = 0;
        for (int a = 0; a < numberString.length(); a++){
            int charAt = numberString.charAt(a) - '0';
            if (charAt == 0 || number % charAt != 0){
                failed++;
            }
        }
        return failed < 1;
    }

}

Edit: Used Mathew Kirschbaum's suggestions to shorten code size.

Answer 62

Mathematica - 93ish characters

Robert G Wilson v gives us the following for Mathematica:

Function definition:

fQ[n_] :=
  Block[{id = Union[IntegerDigits[n]]},
   Union[IntegerQ[#] & /@ (n/id)] == {True}];

To call:

Select[Range[487], fQ[#] &]

Output (ignoring divide by zero complaints):

{1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 
66, 77, 88, 99, 111, 112, 115, 122, 124, 126, 128, 132, 135, 144, 
155, 162, 168, 175, 184, 212, 216, 222, 224, 244, 248, 264, 288, 312, 
315, 324, 333, 336, 366, 384, 396, 412, 424, 432, 444, 448}

Answer 63

Scala (56 chars: 21 for header and 35 for body)

def f(n:Int):Boolean=n==0||n%10!=0&&n%(n%10)==0&&f(n/10)

Old (81):

def f(n:Int):Boolean=if(n==0)true else if(n%10==0)false else n%(n%10)==0&&f(n/10)