25
\$\begingroup\$

To check whether a decimal number is divisible by 7:

Erase the last digit. Multiply it by 2 and subtract from what is left. If the result is divisible by 7, the original number is divisible by 7.

(also described e.g. here)

This rule is good for manual divisibility check. For example:

Is 2016 divisible by 7?

Subtract 6*2 from 201; we get 189. Is this divisible by 7? To check it, let's apply the rule again.

Subtract 9*2 from 18; we get 0. Therefore, 2016 is divisible by 7.

In this challenge, you should apply this rule until the divisibility status is obvious, that is, the number is not greater than 70 (however, see below for details). Make a function or a full program.

Input: a positive integer; your code should support inputs up to 32767 (supporting arbitrary-precision integers is a bonus; see below).

Output: an integer (possibly negative), not greater than 70, that is a result of applying the divisibility-by-7 rule zero or more times.

Test cases:

Input                   Output      Alternative output

1                       1
10                      10          1
100                     10          1
13                      13          -5
42                      42          0
2016                    0
9                       9
99                      -9
9999                    -3
12345                   3
32767                   28          -14

---------- Values below are only relevant for the bonus

700168844221            70          7
36893488147419103232    32          -1
231584178474632390847141970017375815706539969331281128078915168015826259279872    8

Where two possible outputs are specified, either result is correct: the second one corresponds to applying the rule one more time. It's forbidden to apply the rule on a single-digit number: if you erase the digit, nothing (not 0) is left.


Bonus: If your algorithm

where n is the number of decimal digits:

Subtract 50% from your code's byte count.

Real bonus:

In addition, if your algorithm reads the input in normal direction, starting from the most significant digit, subtract 50% once again - your score is 25% of your byte count (it seems possible, but I'm not absolutely sure).

\$\endgroup\$
  • 1
    \$\begingroup\$ @DenkerAffe Returning the input as-is is acceptable. I updated the test-case of input=10 to reflect this; that was the idea from the beginning. \$\endgroup\$ – anatolyg Feb 14 '16 at 20:40
  • 4
    \$\begingroup\$ I wouldn't want to use that rule on 1000000000000000000001. \$\endgroup\$ – Neil Feb 14 '16 at 20:55
  • 1
    \$\begingroup\$ But what if your language has long longs or some equivalent type built in? \$\endgroup\$ – SuperJedi224 Feb 14 '16 at 22:00
  • 1
    \$\begingroup\$ What I was saying was that, in some implementations, it's a 128-bit integer, which is more than big enough for that last test case. \$\endgroup\$ – SuperJedi224 Feb 14 '16 at 22:18
  • 7
    \$\begingroup\$ -1. Not all languages support arbitrary precision. \$\endgroup\$ – March Ho Feb 15 '16 at 3:54

23 Answers 23

23
\$\begingroup\$

Golfscript, 27 22 bytes

{.9>{.10/\10%2*-f}*}:f

You can use it this way:

1000f

Explanation

{.9>{.10/\10%2*-f}*}:f
{                  }:f    # Define block 'f' (similar to a function)
 .                        # Duplicate the first value of the stack
  9>{            }*       # If the value on top of the stack is greater than 9 then the block is executed
     .10/\10%2*-          # Same as nb/10 - (nb%10 * 2) with some stack manipulations '.' to duplicate the top of the stack and '\' to swap the the first and second element of the stack
                f         # Execute block 'f'

5 bytes saved thanks to Dennis !

\$\endgroup\$
  • 1
    \$\begingroup\$ Welcome to Programming Puzzles and Code Golf. This is a good answer, however you could improve it by adding a code breakdown and explanation, like the questions above. To reply to this comment, type @wizzwizz4 (@ then my username) at the beginning of (or anywhere in) a comment. \$\endgroup\$ – wizzwizz4 Feb 14 '16 at 22:05
  • 1
    \$\begingroup\$ @wizzwizz4 Better ? I'm not sure that I understand what you mean by 'code breakdown' (not native speaker sorry) \$\endgroup\$ – Dica Feb 14 '16 at 22:34
  • 8
    \$\begingroup\$ I believe by "code breakdown" he meant an explanation, which you've added. This is a very nice first answer indeed. Welcome to the site! \$\endgroup\$ – Alex A. Feb 14 '16 at 22:55
  • 1
    \$\begingroup\$ You can rewrite the {...}{}if part as {...}*, which will just apply the code block zero of one times, depending on the value pushed by >. Also, we're allowed to perform one more iteration (so replacing 70 with 9 saves a byte), and I don't think you need to pop the block with ;. \$\endgroup\$ – Dennis Feb 15 '16 at 3:12
  • 3
    \$\begingroup\$ @Dica, this is a first answer good enough to get 12+ upvotes on a question with only 624 views, and to get praise from two moderators. If you keep this up, you'll soon overtake Dennis! \$\endgroup\$ – wizzwizz4 Feb 15 '16 at 8:52
13
\$\begingroup\$

Haskell, 35 bytes

until(<71)(\n->div n 10-2*mod n 10)

Usage example: until(<71)(\n->div n 10-2*mod n 10) 36893488147419103232 -> 32.

Nothing much to explain, it's a direct implementation of the algorithm.

\$\endgroup\$
9
\$\begingroup\$

Jelly, 11 bytes

d⁵Uḅ-2µ>9$¿

Try it online!

How it works

d⁵Uḅ-2µ>9$¿  Main link. Input: n

d⁵           Divmod; return [n : 10, n % 10].
  U          Upend; yield [n % 10, n : 10].
   ḅ-2       Convert from base -2 to integer, i.e., yield -2 × (n % 10) + (n : 10).

      µ      Push the previous chain as a link and begin a new, monadic chain.
          ¿  Apply the previous chain while...
       >9$     its return value is greater than 9.
\$\endgroup\$
  • \$\begingroup\$ And as always, Jelly wins. Dennis, how much bytes would it take to implement a jelly interpreter in Jelly? \$\endgroup\$ – Bálint Feb 22 '16 at 12:47
6
\$\begingroup\$

Python 2, 38 bytes

f=lambda x:f(x/10-x%10*2)if x>70else x

Try it here!

Simple recursive approach. Prints x if its < 70 otherwise applies the divisibility rule and calls itself with the result.

\$\endgroup\$
  • \$\begingroup\$ you don't need the space after the ) \$\endgroup\$ – Maltysen Feb 14 '16 at 20:12
  • \$\begingroup\$ @Maltysen True. Copy pasted the wrong one, thanks for the hint! \$\endgroup\$ – Denker Feb 14 '16 at 20:14
  • 2
    \$\begingroup\$ The if is too verbose. f=lambda x:x*(x<70)or f(x/10-x%10*2) \$\endgroup\$ – seequ Feb 14 '16 at 20:25
  • 1
    \$\begingroup\$ @Seeq Nice trick, thanks! This should work in theory, but it reaches the maximum recursion depth with 2016 as input while my version does not. Any idea why? \$\endgroup\$ – Denker Feb 14 '16 at 20:41
  • \$\begingroup\$ Ah, right, didn't consider that. This trick considers x*(x<70) != 0 to be the end condition. If x gets to 0 - like it does with 2016 - the end condition never happens. \$\endgroup\$ – seequ Feb 14 '16 at 20:44
6
\$\begingroup\$

Pyth, 13 bytes

.W>H9-/ZTyeZQ

Try it online: Demonstration or Test Suite

This will print all the alternative answers.

Explanation:

.W>H9-/ZTyeZQ   
            Q   read a number from input
.W              while
  >H9              the number is greater than 9
                do the following with the number:
      /ZT          divide it by 10
     -             and subtract
         yeZ       2*(number%10)
\$\endgroup\$
5
\$\begingroup\$

Julia, 27 26 bytes

f(x)=x>9?f(x÷10-x%10*2):x

This is a recursive function that accepts an integer and returns a BigInt. If the input is a large number like in the last example, Julia parses it as a BigInt, so no manual conversion is necessary.

The approach is just a straightforward implementation of the algorithm. It will produce the alternate outputs. Taking the modulus when dividing by 10 yields the last digit and the quotient from integer division by 10 yields everything but the last digit.

Saved a byte thanks to Dennis!

\$\endgroup\$
  • \$\begingroup\$ We're allowed to perform one more iteration, so replacing 70 with 9 saves a byte. \$\endgroup\$ – Dennis Feb 15 '16 at 3:06
  • \$\begingroup\$ @Dennis Good call, thanks! \$\endgroup\$ – Alex A. Feb 15 '16 at 3:16
4
\$\begingroup\$

Pyth, 17 bytes

L?<b70by-/bT*%bT2

Try it here!

Same recursive approach as in my python answer. Defines a lambda y which is called like this: y12345.
The byte counter in the online interpreter shows 19 bytes because I added the lambda call to it, so you can just try it by hitting the run-button.

Explanation

L?<b70by-/bT*%bT2

L                  # Defines the lambda y with the parameter b
 ?<b70             # if b < 70:
      b            # return b, else:
       -/bT*%bT2   # calculate b/10 - b%10*2 and return it
\$\endgroup\$
  • \$\begingroup\$ You have a typo in your explanation, 17 should be 70 :P \$\endgroup\$ – FryAmTheEggman Feb 15 '16 at 2:01
4
\$\begingroup\$

CJam - 19 bytes

Do-while version:

r~A*{`)]:~~Y*-_9>}g

Try it online or While version #1:

r~{_9>}{`)]:~~Y*-}w

Try it online or While version #2:

r~{_9>}{_A/\A%Y*-}w

Try it online.

r~                     | Read and convert input
  A*                   | Multiply by 10 to get around "if" rule
     `                 | Stringify
      )                | Split last character off
       ]               | Convert stack to array
        :~             | Foreach in array convert to value
          ~            | Dump array
           Y*          | Multiply by 2
             -         | Subtract
              _        | Duplicate
               9>      | Greater than 9?
    {            }g    | do-while
\$\endgroup\$
3
\$\begingroup\$

Oracle SQL 11.2, 116 bytes

WITH v(i)AS(SELECT:1 FROM DUAL UNION ALL SELECT TRUNC(i/10)-(i-TRUNC(i,-1))*2 FROM v WHERE i>70)SELECT MIN(i)FROM v;

Un-golfed

WITH v(i) AS
(
  SELECT :1 FROM DUAL
  UNION ALL
  SELECT TRUNC(i/10)-(i-TRUNC(i,-1))*2 FROM v WHERE i>70
)
SELECT MIN(i) FROM v;
\$\endgroup\$
3
\$\begingroup\$

Haskell, 157 192 184 167 159 147 138+5 bytes - 50% = 71.5 bytes

O(1) space, O(n) time, single-pass!

h d=d%mod d 10
d%r=(quot(r-d)10,r)
p![d]=d-p*10
p![d,e]=d#(e-p)
p!(d:e:f)|(b,a)<-quotRem(2*d)10,(q,r)<-h$e-a-p=(b+q)!(r:f)
m#0=m
m#n=n-2*m
(0!)

Use as 0![6,1,0,2] to apply the rule to 2016, i.e. pass it a number in stream form with least significant figure first. In this way, it will pass over the number digit by digit, applying the rule with O(1) space complexity.

The ungolfed code is here:

import Data.Char

{- sub a b = sub2 0 a b
  where
    sub2 borrow (a:as) (b:bs) = res : sub2 borrow2 as bs
      where
        (borrow2, res) = subDig borrow a b
    sub2 borrow (a:as) [] = sub2 borrow (a:as) (0:[])
    sub2 _ [] _ = [] -}

--subDig :: Int -> Int -> Int -> (Int, Int)
subDig borrow a b = subDig2 (a - b - borrow)
  where
    subDig2 d = subDig3 d (d `mod` 10)
    subDig3 d r = ((r-d) `quot` 10, r)

seven ds = seven2 0 ds
seven2 borrow (d:e:f:gs) = seven2 (b + borrow2) (res:f:gs)
  where
    (a, b) = double d
    (borrow2, res) = subDig borrow e a
seven2 borrow (d:e:[]) = finalApp d (e-borrow)
seven2 borrow (d:[]) = d - borrow*10

double d = ((2*d) `mod` 10, (2*d) `quot` 10)

finalApp m 0 = m
finalApp m n = n - 2*m

num2stream :: Int -> [Int]
num2stream = reverse . map digitToInt . show
sev = seven . num2stream

The gist of how this works is that it implements a digit-by-digit subtraction algorithm, but takes advantage of the fact that each number to be subtracted is at most 2-digits, and so we can subtract an arbitrary amount of these 1-or-2 digit numbers from the main one (as well as eating the least significant digits).

The subtraction algorithm is O(1) and only stores the current 'borrow' value. I altered this to add in the extra digit (either 0 or 1), and we note that this borrow value is bounded (within the range [-2,2] so we need only 3 bits to store this).

The other values stored in memory are temporary variables representing the current 2-digit number to add, a single look-ahead in the stream, and to apply one step of the subtraction algorithm (i.e. it takes two digits and a borrow value, and returns one digit and a new borrow value).

Finally at the end it processes the last two digits in the stream at once to return a single-digit number rather than a list of digits.

N.B. The sev function in the ungolfed version will work on an Integer, converting it into the reversed stream form.

\$\endgroup\$
  • \$\begingroup\$ I intended the bonus to be for the normal order of digits. But I never said it, so it's fair to get the bonus for the reversed order, even though it's less fun. Anyway, even the reversed order is harder than I thought, so it's fun enough! \$\endgroup\$ – anatolyg Feb 15 '16 at 18:24
  • \$\begingroup\$ @anatolyg: Thanks! I'm not sure if it's possible to do a single pass O(1) implementation of the normal order... the rule depends on the least significant figures, so in theory direct application of the rule is impossible except in reversed order. The only other thing I can think of is by finding a mathematically equivalent form - for example Mod[18 - Quotient[n, 10] - 2*n, 21] - 18 + Quotient[n, 10] works empirically for n between 10 and 99, but gets more complicated the more digits n has... \$\endgroup\$ – nitrous Feb 15 '16 at 20:59
  • \$\begingroup\$ Hmm I thought about it and it seemed there might be a way by keeping the front 2 digits and applying each subsequent digit, but multiplying by (-2)^n to take account of it 'filtering through'... as far as I can tell though there's no way to make this work without keeping all the digits in memory and sacrificing the O(1)'ness or even o(n)'ness... I think the normal order is definitely impossible :( \$\endgroup\$ – nitrous Feb 15 '16 at 21:41
  • 1
    \$\begingroup\$ I'm afraid you have to count the bytes of the initial 0 when calling !, too, e.g. as a section (0!) (+ a newline), i.e. +5 bytes. On the other side you can shorten the first to pattern matches of ! to p![d]= and p![d,e]=. Also, use pattern guards instead of the let: p!(d:e:f)|(b,a)<-quotRem(2*d)10,(q,r)<-h$e-a-p=(b+q)!(r:f). \$\endgroup\$ – nimi Feb 16 '16 at 16:55
  • 1
    \$\begingroup\$ @nitrous: Oh I ment (0!) on a line of it's own. (0!) is the function you give as your answer. The 0 is required, but has nothing to do with the input, so you cannot outsource it to the caller. Of course you could also use f x=0!x, but this is longer. \$\endgroup\$ – nimi Feb 16 '16 at 21:32
3
\$\begingroup\$

GNU dc, 20 15 bytes

[10~2*-d70<F]sF

This defines my first (ever) dc function, F. It takes input on the top of stack, and leaves its output at top of stack. Example usage:

36893488147419103232
lFxp
32
\$\endgroup\$
2
\$\begingroup\$

Mathematica, 47 44 bytes

If[#>70,#0[{1,-2}.{⌊#/10⌋,#~Mod~10}],#]&

Simple recursive approach. Could probably be golfed further.

\$\endgroup\$
  • \$\begingroup\$ #0[{1,-2}.QuotientRemainder[#,10]] saves a byte. \$\endgroup\$ – njpipeorgan Feb 15 '16 at 2:42
2
\$\begingroup\$

R, 43 bytes

x=scan();while(x>70)x=floor(x/10)-x%%10*2;x

Explanation:

x=scan()                                      # Takes input as a double
        ;                                     # Next line
         while(x>70)                          # While-loop that runs as long x > 70
                      floor(x/10)             # Divide x by 10 and round that down
                                 -x%%10*2     # Substract twice the last integer
                    x=                        # Update x
                                         ;    # Next line once x <= 70
                                          x   # Print x

Sample runs:

> x=scan();while(x>70)x=floor(x/10)-x%%10*2;x
1: 9999
2: 
Read 1 item
[1] -3

> x=scan();while(x>70)x=floor(x/10)-x%%10*2;x
1: 32767
2: 
Read 1 item
[1] 28
\$\endgroup\$
1
\$\begingroup\$

JavaScript ES6, 38 bytes

a=i=>i>70?a(Math.floor(i/10)-i%10*2):i

Fails with 36893488147419103232 and using ~~(1/10) will also fail for 700168844221

Test:

a=i=>i>70?a(Math.floor(i/10)-i%10*2):i
O.textContent = O.textContent.replace(/(-?\d+) +(-?\d+)/g, (_,i,o) =>
  _+": "+(a(+i)==o?"OK":"Fail")
);
<pre id=O>1                       1
10                      10
100                     10
13                      13
42                      42
2016                    0
9                       9
99                      -9
9999                    -3
12345                   3
700168844221            70
36893488147419103232    32</pre>

\$\endgroup\$
  • \$\begingroup\$ I get two Fails... 70 and 32 \$\endgroup\$ – Conor O'Brien Feb 14 '16 at 20:29
  • \$\begingroup\$ @CᴏɴᴏʀO'Bʀɪᴇɴ Yea me to, I'm still wondering why... \$\endgroup\$ – andlrc Feb 14 '16 at 20:31
  • \$\begingroup\$ Because JavaScript's number type doesn't handle the last case, at least. \$\endgroup\$ – Conor O'Brien Feb 14 '16 at 20:31
  • 1
    \$\begingroup\$ f=n=>n>70?f((n-n%10*21)/10):n is a shorter version but still only works for up to 2**56. \$\endgroup\$ – Neil Feb 14 '16 at 20:37
  • \$\begingroup\$ @Neil see my answer for arbitrary precision, and please feel free to golf, greatly appreciated. \$\endgroup\$ – Patrick Roberts Feb 15 '16 at 5:19
1
\$\begingroup\$

Mathematica, 33 bytes

#//.a_/;a>70:>⌊a/10⌋-2a~Mod~10&

Test case

%[9999]
(* -3 *)
\$\endgroup\$
1
\$\begingroup\$

Perl 5, 47 46 bytes

Had to use bigintfor the last test case. (It returns 20 without)

use bigint;$_=<>;while($_>9){$_-=2*chop;}print

Not really sure it's a candidate for the bonus, so I didn't take it into account. (I think it does, but I'm not really accustomed to the concepts)

Try it here!

\$\endgroup\$
1
\$\begingroup\$

ES6, 108 bytes

f=(s,n=0)=>s>1e9?f(s.slice(0,-1),((1+s.slice(-1)-n%10)%10*21+n-s.slice(-1))/10):s>9?f(((s-=n)-s%10*21)/10):s

Works for 2²⁵⁷ and 1000000000000000000001, but could use further golfing.

\$\endgroup\$
  • \$\begingroup\$ @PatrickRoberts Oops, oversight when reformatting for the submission. \$\endgroup\$ – Neil Feb 15 '16 at 16:25
1
\$\begingroup\$

JavaScript ES6, 140 142 bytes

f=s=>s>9?eval("t=s.replace(/.$/,'-$&*2');for(i=-1;0>(n=eval(u=t[c='slice'](i-4)))&&u!=t;i--);n<0?n:f(t[c](0,i-4)+('0'.repeat(-i)+n)[c](i))"):s

This is true arbitrary-precision math, even works on the largest test-case.

This function recursively removes the last digit from the string, then subtracts 2 * the last digit from the remaining numerical string by iteratively incrementing the amount of digits to apply to the minuend until the difference is positive. Then it appends that difference to the end of the string with appropriately padded 0s and calls itself recursively until its numerical value is less than or equal to 9.

  • Golfed 7 bytes thanks to @Neil (yes I know I gained 2 bytes but I fixed a few bugs that caused the function to freeze or return wrong output for some cases).

f=s=>s>9?eval("t=s.replace(/.$/,'-$&*2');for(i=-1;0>(n=eval(u=t[c='slice'](i-4)))&&u!=t;i--);n<0?n:f(t[c](0,i-4)+('0'.repeat(-i)+n)[c](i))"):s;[['1',1],['10',1],['100',1],['13',-5],['42',0],['2016',0],['9',9],['99',-9],['9999',-3],['12345',3],['700168844221',7],['36893488147419103232',-1],['231584178474632390847141970017375815706539969331281128078915168015826259279872',8]].map(a=>document.write(`<pre>${f(a[0])==a[1]?'PASS':'FAIL'} ${a[0]}=>${a[1]}</pre>`))

\$\endgroup\$
  • \$\begingroup\$ Nice, but it might not work on 1000000000000000000001. \$\endgroup\$ – Neil Feb 15 '16 at 9:29
  • 1
    \$\begingroup\$ Try s.replace(/.$/,'-$&*2'). I don't have any obvious ideas for the rest though sorry. \$\endgroup\$ – Neil Feb 15 '16 at 13:21
1
\$\begingroup\$

C#, 111 104 bytes

int d(int n){var s=""+n;return n<71?n:d(int.Parse(s.Remove(s.Length-1))-int.Parse(""+s[s.Length-1])*2);}
\$\endgroup\$
1
\$\begingroup\$

Brain-Flak, 368 360 bytes

Try it Online!

([([({})]<(())>)](<>)){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}({}<>){{}(({}))(<((()()()()()){}<>)>)<>{({}[()])<>(({}()[({})])){{}(<({}({}))>)}{}<>}{}<>([([([(({}<{}><>)<([{}]{})(<((()()()()()){}(<>))>)<>{({}[()])<>(({}()[({}<({}())>)])){{}(<({}({}<({}[()])>))>)}{}<>}{}<>{}{}({}<>)>){}]{})]<(())>)(<>)]){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}({}<>)}{}

Explanation

To start off all of the code is in a loop that runs until the top of the stack is less than zero:

([([({})]<(())>)](<>)){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}({}<>)
{{}
 ...
 ([([({})]<(())>)](<>)){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}({}<>)
}{}

Inside of the loop we run the divisible by seven algorithm:

Duplicate the top of the stack

(({}))

Take the mod 10 of the top of the stack (last digit)

(<((()()()()()){}<>)>)<>{({}[()])<>(({}()[({})])){{}(<({}({}))>)}{}<>}{}<>({}<{}><>)

This is a bit of a mess but it does the rest of the algorithm I might explain it later but I don't entirely remember how it works:

([(({})<([{}]{})(<((()()()()()){}(<>))>)<>{({}[()])<>(({}()[({}<({}())>)])){{}(<({}({}<({}[()])>))>)}{}<>}{}<>{}{}({}<>)>){}]{})
\$\endgroup\$
1
\$\begingroup\$

C, 56 bytes - 75% = 14

Although this doesn't give the exact same numbers as the test cases, it satisfies the spirit of the question (and arguably more). It correctly identifies exact multiples of 7, and gives the exact remainder for other numbers (since it doesn't ever use negative numbers).

n;f(char*c){for(n=0;*c;)n-=n>6?7:'0'-n-n-*c++;return n;}

There is no multiplication or division in the algorithm, only addition and subtraction, and digits are processed in a single pass from left to right. It works as follows, starting with 0 in the accumulator:

  1. Subtract 7 if necessary, and again if still necessary
  2. Multiply the running total by three, and add the next digit

The "multiply by three" step is written as n-=-n-n to save a byte and to avoid the multiply operator.

When we hit the end, we don't subtract sevens, so the result will be in the range 0-24; if you want a strict modulus (0-7), substitute *c with *c||n>6 in the for loop condition.

It qualifies for the enhanced bonus, because it

  • supports arbitrary-precision integers
  • performs only one pass on the input, in left-to-right order
  • has space complexity O(1)
  • has time complexity O(n).

Test program and results

#include <stdio.h>
int main(int argc, char **argv) {
    while (*++argv)
        printf("%s -> %d\n", *argv, f(*argv));
    return 0;
}
540 -> 15
541 -> 16
542 -> 17
543 -> 18
544 -> 19
545 -> 20
546 -> 21
547 -> 22
548 -> 23
549 -> 24
550 -> 18
99 -> 15
999 -> 12
12345 -> 11
32767 -> 7
700168844221 -> 7
36893488147419103232 -> 11
231584178474632390847141970017375815706539969331281128078915168015826259279872 -> 11

Alternative version

Here's one that recurses (you'll want to enable compiler optimizations to do tail-call transformation or you may overflow your stack; I used gcc -std=c89 -O3):

f(c,n)char*c;{return n>6?f(c,n-7):*c?f(c+1,n+n+n+*c-'0'):n;}

Call it with '0' as the second argument.

Both versions calculate the remainder-modulo-seven of a 60,000 digit number in under 50 milliseconds on my machine.

\$\endgroup\$
  • \$\begingroup\$ Thanks for the bonus - it makes a real change for C to end up so competitive! Currently beaten only by Jelly (11) and Pyth (13). :-) \$\endgroup\$ – Toby Speight Aug 5 '16 at 14:23
1
\$\begingroup\$

PHP, 50 bytes

for($n=$argv[1];$n>9;)$n=$n/10|0-2*($n%10);echo$n;

uses alternative output; works up to PHP_INT_MAX


string version, works for any (positive) number (64 bytes):

for($n=$argv[1];$n>9;)$n=substr($n,0,-1)-2*substr($n,-1);echo$n;
\$\endgroup\$
0
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Java, 133 bytes

int d(int n){String s=""+n;return n<71?n:d(Integer.parseInt(s.replaceFirst(".$",""))-Integer.parseInt(""+s.charAt(s.length()-1))*2);}

I hate how verbose Integer.parseInt is. Ungolfed:

static int div(int n) {
    if (n <= 70) {
        return n;
    } else {
        String num = ("" + n);
        int last = Integer.parseInt("" + num.charAt(num.length() - 1));
        int k = Integer.parseInt(num.replaceFirst(".$", "")) - last * 2;
        return div(k);
    }
}
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