Am I divisible by double the sum of my digits?

Question

Given a positive integer as input, your task is to output a truthy value if the number is divisible by the double of the sum of its digits, and a falsy value otherwise (OEIS A134516). In other words:

(sum_of_digits)*2 | number

• Instead of truthy / falsy values for the true and false cases, you may instead specify any finite set of values for the true/false case, and their complement the other values. For a simple example, you may use 0 for the true case and all other numbers for the false case (or vice versa, if you like).

• Standard input and output rules apply. Default Loopholes also apply.

• You can take input as an integer or as the string representation of that integer.

• This is code-golf, hence the shortest code in bytes wins!

• I am new to PPCG, so I would like you to post an explanation if it's possible.

---

Test Cases

Input - Output - (Reason)

80  - Truthy - (16 divides 80)
100 - Truthy - (2 divides 100)
60  - Truthy - (12 divides 60)
18 - Truthy - (18 divides 18)
12 - Truthy - (6 divides 12)

4 - Falsy - (8 does not divide 4)
8 - Falsy - (16 does not divide 8)
16  - Falsy  - (14 does not divide 16)
21 - Falsy - (6 does not divide 21)
78  - Falsy  - (30 does not divide 78)
110 - Falsy - (4 does not dide 110)
111 - Falsy - (6 does not divide 111)
390 - Falsy  - (24 does not divide 390)

Answer 1

Neim, 3 bytes

ᚫ

Explanation:

      Implicitly convert to int array and sum the digits
 ᚫ     Double
     Is it a divisor of the input?

Try it online!

Detailed version

Answer 2

JavaScript (ES6), 31 29 27 bytes

Takes input as a string. Returns zero for truthy and non-zero for falsy.

n=>n%eval([...n+n].join`+`)

Commented

n => n % eval([...n + n].join`+`)
n =>                                   // take input string n  -> e.g. "80"
                  n + n                // double the input     -> "8080"
              [...     ]               // split                -> ["8", "0", "8", "0"]
                        .join`+`       // join with '+'        -> "8+0+8+0"
         eval(                  )      // evaluate as JS       -> 16
     n %                               // compute n % result   -> 80 % 16 -> 0

Test cases

let f =

n=>n%eval([...n+n].join`+`)

console.log('[Truthy]');
console.log(f("80"))
console.log(f("100"))
console.log(f("60"))
console.log(f("18"))
console.log(f("12"))

console.log('[Falsy]');
console.log(f("4"))
console.log(f("8"))
console.log(f("16"))
console.log(f("21"))
console.log(f("78"))
console.log(f("110"))
console.log(f("111"))
console.log(f("390"))

Answer 3

C#, 46 bytes

using System.Linq;n=>n%(n+"").Sum(c=>c-48)*2<1

Full/Formatted version:

using System;
using System.Linq;

class P
{
    static void Main()
    {
        Func<int, bool> f = n => n % (n + "").Sum(c => c - 48) * 2 < 1;

        Console.WriteLine(f(80));
        Console.WriteLine(f(100));
        Console.WriteLine(f(60));

        Console.WriteLine();

        Console.WriteLine(f(16));
        Console.WriteLine(f(78));
        Console.WriteLine(f(390));

        Console.ReadLine();
    }
}

Answer 4

x86-64 Machine Code, 24 bytes

6A 0A 5E 31 C9 89 F8 99 F7 F6 01 D1 85 C0 75 F7 8D 04 09 99 F7 F7 92 C3

The above code defines a function in 64-bit x86 machine code that determines whether the input value is divisible by double the sum of its digits. The function conforms to the System V AMD64 calling convention, so that it is callable from virtually any language, just as if it were a C function.

It takes a single parameter as input via the EDI register, as per the calling convention, which is the integer to test. (This is assumed to be a positive integer, consistent with the challenge rules, and is required for the CDQ instruction we use to work correctly.)

It returns its result in the EAX register, again, as per the calling convention. The result will be 0 if the input value was divisible by the sum of its digits, and non-zero otherwise. (Basically, an inverse Boolean, exactly like the example given in the challenge rules.)

Its C prototype would be:

int DivisibleByDoubleSumOfDigits(int value);

Here are the ungolfed assembly language instructions, annotated with a brief explanation of the purpose of each instruction:

; EDI == input value
DivisibleByDoubleSumOfDigits:
   push 10
   pop  rsi             ; ESI <= 10
   xor  ecx, ecx        ; ECX <= 0
   mov  eax, edi        ; EAX <= EDI (make copy of input)

SumDigits:
   cdq                  ; EDX <= 0
   div  esi             ; EDX:EAX / 10
   add  ecx, edx        ; ECX += remainder (EDX)
   test eax, eax
   jnz  SumDigits       ; loop while EAX != 0

   lea  eax, [rcx+rcx]  ; EAX <= (ECX * 2)
   cdq                  ; EDX <= 0
   div  edi             ; EDX:EAX / input
   xchg edx, eax        ; put remainder (EDX) in EAX
   ret                  ; return, with result in EAX

In the first block, we do some preliminary initialization of registers:

• PUSH+POP instructions are used as a slow but short way to initialize ESI to 10. This is necessary because the DIV instruction on x86 requires a register operand. (There is no form that divides by an immediate value of, say, 10.)
• XOR is used as a short and fast way to clear the ECX register. This register will serve as the "accumulator" inside of the upcoming loop.
• Finally, a copy of the input value (from EDI) is made, and stored in EAX, which will be clobbered as we go through the loop.

Then, we start looping and summing the digits in the input value. This is based on the x86 DIV instruction, which divides EDX:EAX by its operand, and returns the quotient in EAX and the remainder in EDX. What we'll do here is divide the input value by 10, such that the remainder is the digit in the last place (which we'll add to our accumulator register, ECX), and the quotient is the remaining digits.

• The CDQ instruction is a short way of setting EDX to 0. It actually sign-extends the value in EAX to EDX:EAX, which is what DIV uses as the dividend. We don't actually need sign-extension here, because the input value is unsigned, but CDQ is 1 byte, as opposed to using XOR to clear EDX, which would be 2 bytes.
• Then we DIVide EDX:EAX by ESI (10).
• The remainder (EDX) is added to the accumulator (ECX).
• The EAX register (the quotient) is tested to see if it is equal to 0. If so, we have made it through all of the digits and we fall through. If not, we still have more digits to sum, so we go back to the top of the loop.

Finally, after the loop is finished, we implement number % ((sum_of_digits)*2):

• The LEA instruction is used as a short way to multiply ECX by 2 (or, equivalently, add ECX to itself), and store the result in a different register (in this case, EAX).

  ^{(We could also have done add ecx, ecx+xchg ecx, eax; both are 3 bytes, but the LEA instruction is faster and more typical.)}

• Then, we do a CDQ again to prepare for division. Because EAX will be positive (i.e., unsigned), this has the effect of zeroing EDX, just as before.
• Next is the division, this time dividing EDX:EAX by the input value (an unmolested copy of which still resides in EDI). This is equivalent to modulo, with the remainder in EDX. (The quotient is also put in EAX, but we don't need it.)
• Finally, we XCHG (exchange) the contents of EAX and EDX. Normally, you would do a MOV here, but XCHG is only 1 byte (albeit slower). Because EDX contains the remainder after the division, it will be 0 if the value was evenly divisible or non-zero otherwise. Thus, when we RETurn, EAX (the result) is 0 if the input value was divisible by double the sum of its digits, or non-zero otherwise.

Hopefully that suffices for an explanation.
This isn't the shortest entry, but hey, it looks like it beats almost all of the non-golfing languages! :-)

Answer 5

05AB1E, 5 4 bytes

-1 byte thanks to Okx

SO·Ö

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You can also remove the last Ö to get 0 for truthy and something else for falsy resulting in only 3 bytes but to me that just doesn't seem to appropriately fit the definition.

Explanation

SO·Ö
SO    # Separate the digits and sum them
  ·   # Multiply the result by two
   Ö  # Is the input divisible by the result?

Answer 6

MATL, 7 bytes

tV!UsE\

Outputs 0 if divisible, positive integer otherwise. Specifically, it outputs the remainder of dividing the number by twice the sum of its digits.

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Explanation

t   % Implicit input. Duplicate
V!U % Convert to string, transpose, convert to number: gives column vector of digits
s   % Sum
E   % Times 2
\   % Modulus. Implicit display

Answer 7

Retina, 38 27 bytes

-11 bytes and fixed an error with the code thanks to @MartinEnder

$
$_¶$_
.+$|.
$*
^(.+)¶\1+$

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Prints 1 if divisible, 0 otherwise

Explanation (hope I got this right)

$
$_¶$_

Appends the entire input, plus a newline, plus the input again

.+$|.
$*

Converts each match to unary (either the entire second line which is the original input, or each digit in the first line)

^(.+)¶\1+$

Check if the first line (the doubled digit sum) is a divisor of the second line

Answer 8

Japt, 7 4 bytes

Takes input as a string. Outputs 0 for true or a number greater than 0 for false, which, from other solutions, would appear to be valid. If not, let me know and I'll rollback.

%²¬x

Test it

---

Explanation

Implicit input of string U.
"390"

²

Repeat U twice.
"390390"

¬

Split to array of individual characters.
["3","9","0","3","9","0"]

x

Reduce by summing, automatically casting each character to an integer in the process.
24

%

Get the remainder of dividing U by the result, also automatically casting U to an integer in the process. Implicitly output the resulting integer.
6 (=false)

Answer 9

C89, 55 53 bytes

^{(Thanks to Steadybox!}

s,t;f(x){for(t=x,s=0;t;t/=10)s+=t%10;return x%(s*2);}

It takes a single input, x, which is the value to test. It returns 0 if x is evenly divisible by double the sum of its digits, or non-zero otherwise.

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Ungolfed:

/* int */ s, t;
/*int */ f(/* int */ x)
{
    for (t = x, s = 0; t /* != 0 */; t /= 10)
        s += (t % 10);
    return x % (s * 2);
}

As you can see, this takes advantage of C89's implicit-int rules. The global variables s and t are implicitly declared as ints. (They're also implicitly initialized to 0 because they are globals, but we can't take advantage of this if we want the function to be callable multiple times.)

Similarly, the function, f, takes a single parameter, x, which is implicitly an int, and it returns an int.

The code inside of the function is fairly straightforward, although the for loop will look awfully strange if you're unfamiliar with the syntax. Basically, a for loop header in C contains three parts:

for (initialization; loop condition; increment)

In the "initialization" section, we've initialized our global variables. This will run once, before the loop is entered.

In the "loop condition" section, we've specified on what condition the loop should continue. This much should be obvious.

In the "increment" section, we've basically put arbitrary code, since this will run at the end of every loop.

The larger purpose of the loop is to iterate through each digit in the input value, adding them to s. Finally, after the loop has finished, s is doubled and taken modulo x to see if it is evenly divisible. (A better, more detailed explanation of the logic here can be found in my other answer, on which this one is based.)

Human-readable version:

int f(int x)
{
    int temp = x;
    int sum  = 0;
    while (temp > 0)
    {
        sum  += temp % 10;
        temp /= 10;
    }
    return x % (sum * 2);
}

Answer 10

Brachylog, 8 bytes

ẹ+×₂;I×?

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Explanation

ẹ+           Sum the digits
  ×₂         Double
    ;I×?     There is an integer I such that I×(double of the sum) = Input

Answer 11

Python 2, 34 32 bytes

-2 bytes thanks to @Rod

lambda n:n%sum(map(int,`n`)*2)<1

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Answer 12

Mathematica, 26 bytes

(2Tr@IntegerDigits@#)∣#&

No clue why ∣ has a higher precedence than multiplication...

Answer 13

PHP, 41 bytes

prints zero if divisible, positive integer otherwise.

<?=$argn%(2*array_sum(str_split($argn)));

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Answer 14

Excel, 63 bytes

=MOD(A1,2*SUMPRODUCT(--MID(A1,ROW(OFFSET(A$1,,,LEN(A1))),1)))=0

Summing digits is the lengthy bit.

Answer 15

Perl 6, 19 bytes

{$_%%(2*.comb.sum)}

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Answer 16

Husk, 9 8 bytes

Thanks to Leo for saving 1 byte.

Ṡ¦ȯ*2ṁis

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Explanation

Ṡ¦ȯ         Test whether f(x) divides x, where f is the function obtained by
            composing the next 4 functions.
       s    Convert x to a string.
     ṁi     Convert each character to an integer and sum the result.
   *2       Double the result.

Answer 17

Haskell, 38 37 42 bytes

Thanks to Zgarb for golfing off 1 byte

f x=read x`mod`foldr((+).(*2).read.pure)0x

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Takes input as a string; returns 0 if divisible and nonzero otherwise.

Answer 18

TI-BASIC, 27 26 21 bytes

-5 thanks to @Oki

:fPart(Ans/sum(2int(10fPart(Ans10^(~randIntNoRep(1,1+int(log(Ans

This is made trickier by the fact that there is no concise way to sum integer digits in TI-BASIC. Returns 0 for True, and a different number for False.

Explanation:

:fPart(Ans/sum(2int(10fPart(Ans10^(-randIntNoRep(1,1+int(log(Ans
                               10^(-randIntNoRep(1,1+int(log(Ans #Create a list of negative powers of ten, based off the length of the input, i.e. {1,0.1,0.01}
                            Ans                                  #Scalar multiply the input into the list
                    10fPart(                                     #Remove everything left of the decimal point and multiply by 10
               2int(                                             #Remove everything right of the decimal point and multiply by 2
           sum(                                                  #Sum the resulting list
       Ans/                                                      #Divide the input by the sum
:fPart(                                                          #Remove everything left of the decimal, implicit print

Answer 19

Python 3, 35 bytes

lambda a:a%(sum(map(int,str(a)))*2)

Answer 20

Braingolf, 13 12 bytes

VR.Mvd&+2*c%

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Outputs 0 for truthy, any other number for falsey.

Explanation

VR.Mvd&+2*c%  Implicit input from command-line args
VR            Create stack2, return to stack1
  .M          Duplicate input to stack2
    vd        Switch to stack2, split into digits
      &+      Sum up all digits
        2*    Double
          c   Collapse stack2 back into stack1
           %  Modulus
              Implicit output of last item on stack

Answer 21

Java, 66 bytes

-1 byte thanks to Olivier

a->{int i=0;for(int b:(""+a).getBytes())i+=b-48;return a%(i*2)<1;}

Ungolfed & explanation:

a -> {
    int i = 0;

    for(int b : (""+a).getBytes()) { // Loop through each byte of the input converted to a string
        i += b-48; // Subtract 48 from the byte and add it to i
    }

    return a % (i*2) < 1 // Check if a % (i*2) is equal to one
    // I use <1 here for golfing, as the result of a modulus operation should never be less than 0
}

Answer 22

Japt, 7 bytes

vUì x*2

Returns 1 for true, 0 for false

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Explanation

vUì x*2
v        // Return 1 if the input is divisible by:
 Uì      //   Input split into a base-10 array
    x    //   Sum the array
     *2  //   While mapped by *2

Answer 23

Haskell, 49 Bytes

f x=(==) 0.mod x.(*)2.sum.map(read.return).show$x

Usage

f 80

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Answer 24

J, 15 bytes

0 indicates truthy, nonzero indicates falsy.

|~[:+/2#"."0@":

Explanation

        "."0@":  convert to list of digits
  [:+/2#         sum 2 copies of the list ([: forces monadic phrase)
|~               residue of sum divided by argument?

Answer 25

Ohm, 5 bytes

D}Σd¥

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Answer 26

tcl, 45

puts [expr 1>$n%(2*([join [split $n ""] +]))]

demo

Answer 27

Haskell, 35 34 bytes

f x=mod x$2*sum[read[c]|c<-show x]

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Returns '0' in the true case, the remainder otherwise.

Haskell, pointfree edition by nimi, 34 bytes

mod<*>(2*).sum.map(read.pure).show

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Answer 28

Mini-Flak, 296 292 bytes

({}((()[()]))){(({}[((((()()()){}){}){}){}])({}(((({})){}){}{}){})[({})])(({}({})))({}(({}[{}]))[{}])({}[({})](({}{})(({}({})))({}(({}[{}]))[{}])({}[({})]))[{}])(({}({})))({}(({}[{}]))[{}])({}[({})])}{}({}(((({}){})))[{}]){({}((({}[()]))([{(({})[{}])()}{}]()){{}{}(({}))(()[()])}{})[{}][()])}

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The TIO link have more comments from me, so it is partially easier to read.

Truthy/Falsey: Truthy (divisible) if the second number is equal to the third number, falsy otherwise. So both the truthy and falsy set are infinite, but I suppose that should be allowed. +10 byte if that is not.

Note: Leading/trailing newlines/whitespaces are not allowed in input.

Answer 29

Java (OpenJDK 8), 55 53 bytes

a->{int i=0,x=a;for(;x>0;x/=10)i+=x%10*2;return a%i;}

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A return value of 0 means truthy, anything else means falsy.

Since my comment in Okx's answer made no ripple, I deleted it and posted it as this answer, golfed even a bit more.

Further golfing thanks to @KrzysztofCichocki and @Laikoni who rightfully showed me I needn't answer a truthy/falsy value, but any value as long as I describe the result.

Answer 30

PHP, 44 bytes

for(;~$d=$argn[$i++];)$t+=2*$d;echo$argn%$t;

Run like this:

echo 80 | php -nR 'for(;~$d=$argn[$i++];)$t+=2*$d;echo$argn%$t;'

Explanation

Iterates over the digits to compute the total, then outputs the modulo like most answers.

Answer 31

APL, 13 bytes

{⍵|⍨2×+/⍎¨⍕⍵}

Outputs 0 as a truthy value, and any other as falsy.

Answer 32

C# (.NET Core), 51 bytes

n=>{int l=n,k=0;for(;l>0;l/=10)k+=l%10;return n%k;}

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0 for truthy, anything else for falsey.

Answer 33

Perl 5, 21 + 1 (-p) = 22 bytes

$_%=eval s/./+2*$&/rg

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Answer 34

Pyth - 12 Bytes

%sZ*2ssMscZ1

Explanation:

%       - implicitly print modulus of
 s      - parse int
  Z     - input
 *      - two times
  s    - sum of
   M   - map
    s  - parse int to
    c  - substrings
     Z - of input
     1 - of length 1

Answer 35

Befunge-93, 34 bytes

p&::v>0g2*%.@
+19:_^#:/+91p00+g00%

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Prints 0 for truthy, a different integer for falsey.

Answer 36

Brachylog, 7 bytes

f.&ẹj+∈

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The predicate succeeds if the input is divisible by twice the sum of its digits, and fails otherwise. Run as a program, success prints true. and failure prints false..

This started off as a minor edit to Fatalize's answer, golfing a byte off by replacing +×₂ with j+, but as I was testing it I noticed that one or another of the updates he's made to the language in the last year and a half seems to have broken his solution on the case of 18, with I apparently being unable to assume the value of 1, so I ended up having to restructure it somewhat (fortunately without changing the byte count).

f.         The output variable is a list of the input's factors.
  &        And,
   ẹ       the digits of the input,
    j      all duplicated,
     +     sum up to
      ∈    a member of the output variable.

Answer 37

K (ngn/k), 15 14 bytes

{(2*+/10\x)!x}

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---

0 is truthy

Answer 38

Keg+Reg, 11 bytes

Boring algorithm.

¿:;9%1+2*%

Finds the digital root of the top of the stack and then finds the remainder.

0 is a truthy value, any value that is not 0 is a falsy value. (Including ASCII characters)

Explanation (Syntactically invalid)

¿#         Integer input
 :#        Duplicate
  ;9%1+#   Digital Root Formula (1+(n-1)mod 9)
       2*%#Multiply by 2 and find the remainder

I believe everyone here used repetition/reduction but not an existing formula.

Answer 39

Jelly, 4 bytes

DSḤḍ

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Explanation:

DSḤḍ Takes an integer as input.
D    Get digits (convert to base 10)
 S   Sum
  Ḥ  Unhalve (double)
   ḍ Check if y (the input itself) is divisible by x

Answer 40

Pari/GP, 24 bytes

n->!(n%(2*sumdigits(n)))

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Answer 41

jq, 37 characters

.%(tostring/""|map(tonumber)|add*2)<1

Sample run:

bash-4.4$ jq '.%(tostring/""|map(tonumber)|add*2)<1' <<< '60'
true

bash-4.4$ jq '.%(tostring/""|map(tonumber)|add*2)<1' <<< '390'
false

Try on jp‣play

Answer 42

PowerShell, 39 bytes

($a="$args")%($a*2-replace'\B','+'|iex)

0 as Truthy output, any other number as Falsey.

PS D:\> D:\t.ps1 80
0

PS D:\> D:\t.ps1 81
9

It takes the args array as a string, does string multiplication to double all the digits, regex-replaces non-word-boundaries to make '8+0+8+0', eval's that to get the sum, and calculates a remainder of original/sum.

Answer 43

CJam, 11 10 bytes

1 byte saved thanks to @Challenger5

{_Ab:+2*%}

Anonymous block that takes the input from the stack and replaces it by the output, which is 0 if divisible or a positive integer otherwise.

Try it online! Or verify all test cases.

Explanation

{        }   e# Block
 _           e# Duplicate
  Ab         e# Convert to base 10. Gives an array of digits
    :+       e# Fold addition over array
      2*     e# Multiply by 2
        %    e# Modulus

Answer 44

Bash + coreutils + bc, 46 bytes

yes $1%\(0`echo $1|sed 's/\(.\)/+\1*2/g'`\)|bc

Input via command line. 0 is truthy.

First I tried to loop through the characters of the input string, but bash loops are to big. I ended up using sed to replace each character with an addition and multiplication, which then goes to bc. I saved a byte using yes instead of echo.

Answer 45

Perl 5, 27 bytes

$_%(2*eval join'+',split//)

Returns 0 for Truthy and non-zero for Falsy

Answer 46

Ruby, 25 bytes

->x{x%(x.digits.sum*2)<1}

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Answer 47

LOGO, 24 bytes

[modulo ? 2*reduce "+ ?]

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Only FMSLogo and its older version (MSWLogo) support ? in template-list; and only FMSLogo supports apply for infix operator +.

That is a template-list (equivalent of lambda function in other languages). Return 0 for truthy and other values for falsy.

Usage:

pr invoke [modulo ? 2*reduce "+ ?] 100

(invoke calls the function, pr prints the result)

Explanation:

There is only one points need explaining: reduce. That is similar to fold in some other languages: reduce "f [1 2 3 4] calculate (f (f (f 1 2) 3) 4). In this case, reduce "+ ? calculate total of elements of digits in ?.

Numbers in LOGO is represented by word, so reduce, a library function can receive a word as its second parameter (while apply can't)

Answer 48

Alice, 16 bytes

/o. &
\i@h /+.+F

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Prints twice the digit sum as the truthy result and 0 otherwise.

The code looks so painfully suboptimal with the two spaces and the repeated . and +, but I just can't figure out a more compact layout.

Explanation

/    Switch to Ordinal mode.
i    Read all input as a string.
.    Duplicate.
h    Split off the first character.
&    Fold the next command over the remaining characters (i.e. push each character
     in turn and then execute the command).
/    Switch back to Cardinal mode (not a command).
+    Add. This is folded over the characters, which implicitly converts them 
     to the integer value they represent and adds them to the leading digit.
.+   Duplicate and add, to double the digit sum.
F    Test for divisibility. Gives 0 if the input is not divisible by twice its
     digit sum and the (positive/truthy) divisor otherwise.
\    Switch to Ordinal mode.
o    Implicitly convert the result to a string and print it.
@    Terminate the program.

Answer 49

Java 7, 197 133 126 bytes

First try Second try, fixed it, -9 characters on variable name, and changing it into function for the other subtraction with help of Olivier Grégoire

Changed boolean to int, now return 0 for true values, other number for falsy

int b(String a){int t=0;for(int i=0;i<a.length();i++){t+=Integer.parseInt(""+a.charAt(i));}return(Integer.parseInt(a)%(t*2));}

boolean a(String a){int t=0;for(int i=0;i<a.length();i++){t+=Integer.parseInt(""+a.charAt(i));}return(Integer.parseInt(a)%(t*2)==0);}

Ungolfed

int b(String a){
int t=0;for(int i=0;i<a.length();i++){
t+=Integer.parseInt(""+a.charAt(i));
}
return(Integer.parseInt(a)%(t*2));}

Answer 50

Pyth, 7 bytes

%QyssM`

Test it online! It returns 0 for truthy, a non null positive integer for falsy.

Explanations

       Q    # Implicit input Q
      `     # Convert the input to a string
    sM      # For each character in Q, convert to integer
   s        # Sum the resulting list
  y         # Multiply by two
%Q          # Perform the modulo with the input

Answer 51

Java, 53 bytes

a->{int i=0,c=a;for(;c>0;c/=10)i+=c%10*2;return a%i;}

0 true

>0 false

Answer 52

Python, 150 characters

def func(number)
    num = number #your number
    sum = 0
    while(num!=0):
        sum += num%10
        num /= 10
    return (2*sum) % number == 0

Answer 53

R, 71 bytes

f=function(x)`if`(x%%(sum(floor(x/10^(0:(nchar(x)-1)))%%10)*2)==0,T,F)

Using floor(x/10^(0:(nchar(x)-1)))%%10 saves the usual hassle of converting to character and back.

> f(1)
[1] FALSE
> f(12)
[1] TRUE
> f(123456789)
[1] FALSE

Answer 54

R, 58 bytes

(a<-scan())%%(2*sum(strtoi(el(strsplit(paste(a),"")))))<1

a <- scan() stores user's input in variable a.

This variable is of type numeric. The paste function converts it into character for the strsplit function to separate its digits. el permits to use only the first element of the list created by strsplit.

The strtoi function converts then the digits back into numeric, for the function sum to, well, sum, before doubling the result.

To finish, divisibility of a with the double sum of its digits is tested by the ... %% ... < 1 part, outputing TRUE or FALSE.

Answer 55

><>, 26 bytes

000\~2*%n;
0(?\c%:{a*+}+i:

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Output is 0 for truthy, non-0 for falsey.

Answer 56

Excel VBA, 72 Bytes

Anonymous VBE immediate window function that takes input from cell [A1] and outputs to the VBE immediate window

For i=1To[Len(A1)]:s=s+Int(Mid([A1],i,1)):Next:s=s*2:?Int([A1]/s)=[A1]/s

Answer 57

Element, 29 bytes

Here goes my first "real" codegolf attempt...

_2:x;2:$'0 z;[(z~+z;]x~z~2*%`

Outputs "0" if true, anything else if false.

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Explanation:

_             push input into main stack
2:            pop number and push it twice
x;            store one of them as x
2:            pop again and push twice 
$'            pop and get length, push it to control stack
0 z;          initiate accumulator "z" with value zero 
[             for length of input number (loop start), 
(             pop off first digit,
z~            retrieve "z",
+             add popped digit to it,
z;            store "z"  again 
]             (end of looped statement)
x~            retrieve "x" (copy of original number)
z~            retrieve "z" (accumulated digit sum)
2*            double accumulated digit sum
%             modulus the digit sum and original number
`             pop result to output.

Answer 58

Java (OpenJDK 8), 40 bytes

i->i%(i+"").chars().map(k->k*2-96).sum()

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Returns 0 as truthy value and a value greater than 0 as falsy value.

Answer 59

Julia, 30 29 bytes

f(x)=x/2sum(digits(x)) in 0:x

saved a byte thanks to caird-coinheringaahing

Answer 60

K (oK), 14 bytes

Solution:

(2*+/48!$x)!x:

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Examples:

  (2*+/48!$x)!x:18
0
  (2*+/48!$x)!x:123
3
  (2*+/48!$x)!x:390
6

Explanation:

Evaluation is performed right-to-left, unfortunately modulo is b!a rather than a!b hence the need for brackets. Returns 0 for truthy, positive integer for falsey per spec.

(2*+/48!$x)!x: / the solution
            x: / store input in variable x
           !   / modulo right by the left ( e.g. 3!10 => 1, aka 10 mod 3 ) 
(         )    / the left
        $x     / string ($) input, 123 => "123"
     48!       / modulo with 48, "123" => 1 2 3
   +/          / sum over, +/1 2 3 => 6
 2*            / double, 6 => 12

Notes:

• Prepend solution with ~ (not) to give a true/false result.

Answer 61

Add++, 11 bytes

L,EDEs;A@%!

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First time I've used either a Lambda or ; in an Add++ answer

How it works

L,   - Create a lambda function. Example argument; 172
  ED - Push the digits;   STACK = [[1 7 2]]
  Es - Stack-clean sum;   STACK = [10]
  ;  - Double;            STACK = [20]
  A  - Push the argument; STACK = [20 172] 
  @% - Modulo;            STACK = [12]
  !  - Logical NOT;       STACK = [0]

Answer 62

><>, 36 Bytes

Unlike the other ><> response, this takes the number itself as input as opposed to the digits. Of course, because of that it's a bit longer longer.

:&>:a%:}-\1-?\2*&$%n;
  \?)0:,a/l +<

Explanation:

:&

Duplicates input, stores a copy in the register.

  >:a%:}-\
  \?)0:,a/

Converts the number into its digits.

         \1-?\
         /l +<

Sum the digits

              2*&$%n;

Multiplies the digit sum by 2, and prints the original number modulo that sum. Output is, then, 0 for truthy and something else otherwise.

Answer 63

Ly, 10 bytes

nsS&+2*lf%

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Ly is lucky enough to have a splitting built-in.

Returns 0 for truthy and a number greater than 0 for falsy.

Answer 64

Deorst, 16 bytes

iE"E|miEH:+zE|%N

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How it works

Implicitly inputs and outputs, example: 172

iE"              - To string;  STACK = ['172']
   E|            - Splat;      STACK = ['1' '7' '2']
     mi          - To integer; STACK = [1 7 2]
       EH        - Sum;        STACK = [10]
         :+      - Double;     STACK = [20]
           zE|   - Push input; STACK = [20 172]
              %N - Divides?    STACK = [0]

Answer 65

ARBLE, 33 bytes

nt(a%(sum(explode(""..a,"."))*2))

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Answer 66

Clojure, 59 bytes

(fn[n](= 0(rem n(* 2(apply +(map #(-(int %)48)(str n)))))))

---

(defn sum-of-digits? [n]
  (let [ds (map #(- (int %) 48) (str n)) ; Stringify n, and parse each digit char of the string
        d-sum (* 2 (apply + ds))] ; Get the doubled sum
    (= 0 (rem n d-sum)))) ; Check if the remainder of (/ n sum) is 0

Answer 67

Runic Enchantments, 11 bytes

i:'+A2*%0=@

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Explanation

i:             Read input, duplicate it
  '+A          Sum its digits
     2*        Multiply by 2
       %       Modulo input with digit sum
        0=     Compare to 0
          @    Print and terminate

Answer 68

Forth (gforth), 58 bytes

: f 0 over begin 10 /mod >r + r> ?dup 0= until 2* mod 0= ;

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Code Explanation

: f           \ start a new word definition
  0 over      \ create an accumulator and copy the input above it on the stack
  begin       \ start an indefinite loop
    10 /mod   \ divide by 10 and get quotient and remainder
    >r + r>   \ add remainder to accumulator
    ?dup      \ duplicate the quotient if it's greater than 0
    0=        \ check if quotient is 0
  until       \ end the indefinite loop if it is
  2* mod      \ multiply result by 2, then modulo with input number
  0=          \ check if result is 0 (input is divisible by result)
;             \ end word definition