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.

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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

Ohm, 5 bytes

D}Σd¥

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Answer 2

tcl, 45

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

demo

Answer 3

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 4

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 5

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 6

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 7

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 8

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 9

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

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

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Answer 10

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 11

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 12

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 13

Befunge-93, 34 bytes

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

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Prints 0 for truthy, a different integer for falsey.

Answer 14

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 15

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

Answer 16

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 17

Vyxal, 3 bytes

∑dœ

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Compiles as:

stack.push(summate(stack.pop()))
stack.push(stack.pop() * 2)
lhs, rhs = stack.pop(2); stack.push(int((rhs % lhs) == 0))

Explained

∑dœ
∑   # Push the sum of the integers of the input
 d  # Double that sum
  œ # And push ((sum(input) * 2) % input) == 0

Answer 18

MathGolf, 3 bytes

Σ∞÷

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Explanation:

Σ    # Get the digit-sum of the (implicit) input-integer
 ∞   # Double it
  ÷  # Check if the (implicit) input-integer is divisible by this value
     # (after which the entire stack is output implicitly as result)

Answer 19

Whitespace, 103 bytes

[S S S N
_Push_0][S N
S _Duplicate_0][S N
S _Duplicate_0][T   N
T   T   _Read_STDIN_as_integer][T   T   T   _Retrieve_input][N
S S N
_Create_Label_LOOP][S N
S _Duplicate][N
T   S S N
_If_0_Jump_to_Label_DONE][S N
S _Duplicate][S S S T   S T S N
_Push_10][T S T T   _Modulo][S T    S S T   S N
_Copy_0-based_2nd][T    S S S _Add][S N
T   _Swap_top_two][S S S T  S T S N
_Push_10][T S T S _Integer_divide][N
S N
N
_Jump_to_Label_LOOP][N
S S S N
_Create_Label_DONE][T   T   T   _Retrieve_input][S N
T   _Swap_top_two][S S S T  S N
_Push_2][T  S S N
_Multiply]{T    S T T   _Modulo][T  N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Outputs 0 as truthy or a positive integer as falsey.

Try it online (with raw spaces, tabs and new-lines only).

To output two distinct values (0 for truthy; 1 for falsey) it would be 128 bytes instead: try it online.

Explanation in pseudo-code:

Integer sum = 0
Integer input = STDIN as integer
Integer n = input
Start LOOP:
  If(n == 0):
    Jump to Label DONE
  Integer t = n modulo-10
  sum = sum + t
  n = n integer-divided by 10
  Go to next iteration of LOOP

DONE:
  sum = sum * 2
  Integer result = input % sum
  Print result as integer to STDOUT

Answer 20

Factor + math.text.utils math.unicode, 37 bytes

[ dup 1 digit-groups Σ 2 * mod 0 = ]

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Explanation

                ! 12
dup             ! 12 12
1 digit-groups  ! 12 { 2 1 }
Σ               ! 12 3
2 *             ! 12 6
mod             ! 0
0 =             ! t

Answer 21

Vyxal, 3 bytes

∑dḊ

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Answer 22

C++, 121 bytes

#include<iostream>
main(){std::string n;int m=0;std::cin>>n;for(char i:n){m+=i-48;};std::cout<<((stoi(n)%(m*2)==0)?0:1);}

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Explanation

#include<iostream>
main(){
  std::string n; int m = 0; // Initialize `n` (input) and `m` (sum)
  std::cin >> n; // Get input
  for(char i:n) // Loop through input to calculate the sum of digits
  {
    m+=i-48;
  };
  std::cout<<((stoi(n)%(m*2)==0)?0:1); // Output `0` if double the sum of digits is divisible by the number, else return `1`
}

Answer 23

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 24

Pari/GP, 24 bytes

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

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Answer 25

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 26

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 27

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 28

Perl 5, 27 bytes

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

Returns 0 for Truthy and non-zero for Falsy

Answer 29

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 30

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)