Divide a number by 3 without using - operators

Question

How would you divide a number by 3 without using          -      operators   The number may be signed or unsigned

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if we consider   div   is not orthographically    def divBy3 n       return n   div   3   print divBy3 9    or   9  3

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How about this approach  c     private int dividedBy3 int n            List lt Object gt  a   new Object n  ToList            List lt Object gt  b   new List lt object gt             while  a Count  gt  2                a RemoveRange 0  3               b Add new Object                       return b Count

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All answers are probably not that what the interviewer liked to hear   My answer       I would never do that  who will me pay for such silly things  Nobody   will have an advantage on that  its not faster  its only silly    Prozessor designers have to know that  but this must then work for all numbers  not only for division by 3

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Yet another solution  This should handle all ints  including negative ints  except the min value of an int  which would need to be handled as a hard coded exception  This basically does division by subtraction but only using bit operators  shifts  xor   amp  and complement   For faster speed  it subtracts 3    decreasing powers of 2   In c   it executes around 444 of these DivideBy3 calls per millisecond  2 2 seconds for 1 000 000 divides   so not horrendously slow  but no where near as fast as a simple x 3  By comparison  Coodey s nice solution is about 5 times faster than this one   public static int DivideBy3 int a        bool negative   a  lt  0      if  negative  a   Negate a       int result      int sub   3  lt  lt  29      int threes   1  lt  lt  29      result   0      while  threes  gt  0            if  a  gt   sub                a   Add a  Negate sub                result   Add result  threes                     sub  gt  gt   1          threes  gt  gt   1            if  negative  result   Negate result       return result    public static int Negate int a        return Add  a  1     public static int Add int a  int b        int x   0      x   a   b      while   a  amp  b     0            b    a  amp  b   lt  lt  1          a   x          x   a   b            return x      This is c  because that s what I had handy  but differences from c should be minor

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This should work for any divisor  not only three  Currently only for unsigned  but extending it to signed should not be that difficult    include  lt stdio h gt   unsigned sub unsigned two  unsigned one   unsigned bitdiv unsigned top  unsigned bot   unsigned sub unsigned two  unsigned one    unsigned bor  bor   one  do                one    two  amp  bor          two    bor          bor   one lt  lt 1            while  one   return two     unsigned bitdiv unsigned top  unsigned bot    unsigned result  shift   if   bot    top  lt  bot  return 0   for shift 1 top  gt    bot lt  lt  1   shift        bot  gt  gt   1   for  result 0  shift--  bot  gt  gt   1             result  lt  lt  1          if  top  gt   bot                    top   sub top bot                   result    1                              return result     int main void    unsigned arg val   for  arg 2  arg  lt  40  arg              val   bitdiv arg 3           printf  Arg  u Val  u n   arg  val             return 0

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Use cblas  included as part of OS X s Accelerate framework     02 31 59   william relativity     cat div3 c  import  lt stdio h gt   import  lt Accelerate Accelerate h gt   int main         float multiplicand   123456 0      float multiplier   0 333333      printf   f    f       multiplicand  multiplier       cblas sscal 1  multiplier   amp multiplicand  1       printf   f n   multiplicand       02 32 07   william relativity     clang div3 c -framework Accelerate -o div3  amp  amp    div3 123456 000000   0 333333    41151 957031

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Why don t we just apply the definition studied at College  The result maybe inefficient but clear  as the multiplication is just a recursive subtraction and subtraction is an addition  then the addition can be performed by a recursive xor and logic port combination    include  lt stdio h gt   int add int a  int b      int rc     int carry     rc   a   b      carry    a  amp  b   lt  lt  1     if  rc  amp  carry         return add rc  carry      else       return rc   carry      int sub int a  int b      return add a  add  b  1        int div  int D  int Q        lets do only positive and then    add the sign at the end    inversion needs to be performed only for  Q -D or -Q  D        int result 0     int sign 0     if  D  lt  0           D sub 0 D         if  Q lt 0            Q sub 0 Q         else          sign 1       else         if  Q lt 0              Q sub 0 Q            sign 1                   while D gt  Q          D   sub  D  Q          result              Apply sign       if  sign         result   sub 0 result      return result     int main  int argc  char    argv          printf   2 plus 3  d n   add 2 3         printf   22 div 3  d n   div 22 3         printf   -22 div 3  d n   div -22 3         printf   -22 div -3  d n   div -22 -3         printf   22 div 03  d n   div 22 -3         return 0      As someone says    first make this work  Note that algorithm should work for negative Q

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Using counters is a basic solution   int DivBy3 int num        int result   0      int counter   0      while  1            if  num    counter          Modulus 0             return result          counter   abs  counter        counter          if  num    counter          Modulus 1             return result          counter   abs  counter        counter          if  num    counter          Modulus 2             return result          counter   abs  counter        counter          result   abs  result          result           It is also easy to perform a modulus function  check the comments

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Would it be cheating to use the   operator  behind the scenes  by using eval and string concatenation   For example  in Javacript  you can do  function div3  n        var div   String fromCharCode 47       return eval  n  div  3  join

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Write the program in Pascal and use the DIV operator   Since the question is tagged c  you can probably write a function in Pascal and call it from your C program  the method for doing so is system-specific   But here s an example that works on my Ubuntu system with the Free Pascal fp-compiler package installed    I m doing this out of sheer misplaced stubbornness  I make no claim that this is useful    divide by 3 pas    unit Divide By 3  interface     function div by 3 n  integer   integer  cdecl  export  implementation     function div by 3 n  integer   integer  cdecl      begin         div by 3    n div 3      end  end    main c     include  lt stdio h gt   include  lt stdlib h gt   extern int div by 3 int n    int main void        int n      fputs  Enter a number     stdout       fflush stdout       scanf   d    amp n       printf   d   3    d n   n  div by 3 n        return 0      To build   fpc divide by 3 pas  amp  amp  gcc divide by 3 o main c -o main   Sample execution       main Enter a number  100 100   3   33

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Idiotic conditions call for an idiotic solution    include  lt stdio h gt   include  lt stdlib h gt   int main         FILE   fp fopen  temp dat   w b        int number 12346      int divisor 3      char   buf   calloc number 1       fwrite buf number 1 fp       rewind fp       int result fread buf divisor number fp       printf   d    d    d   number  divisor  result       free buf       fclose fp       return 0      If also the decimal part is needed  just declare result as double and add to it the result of fmod number divisor    Explanation of how it works   The fwrite writes number bytes  number being 123456 in the example above   rewind resets the file pointer to the front of the file  fread reads a maximum of number  records  that are divisor in length from the file  and returns the number of elements it read    If you write 30 bytes then read back the file in units of 3  you get 10  units    30   3   10

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Using BC Math in PHP    lt  php      a   12345       b   bcdiv  a  3        gt      MySQL  it s an interview from Oracle    gt  SELECT 12345 DIV 3      Pascal   a   12345  b   a div 3      x86-64 assembly language   mov  r8  3 xor  rdx  rdx    mov  rax  12345 idiv r8

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Here s a method my Grandfather taught me when I was a child  It requires   and   operators but it makes calculations easy   Add the individual digits together and then see if its a multiple of 3   But this method works for numbers above 12   Example  36   3 6 9 which is a multiple of 3   42   4 2 6 which is a multiple of 3

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Here it is in Python with  basically  string comparisons and a state machine   def divide by 3 input     to do        enque index   0   zero to 9    0  1  2  3  4  5  6  7  8  9    leave over   0   for left over in  0  1  2       for digit in zero to 9          left over  digit   gt  enque  leave over       to do  left over  digit      zero to 9 enque index   leave over        if leave over    0          leave over   1       elif leave over    1          leave over   2       elif leave over    2 and enque index    9          leave over   0         enque index    1  2  3  4  5  6  7  8  9  enque index    answer q        left over   0   digits   list str input     if digits 0      -       answer q append  -     digits   digits 1     for digit in digits      enque  left over   to do  left over  int digit        if enque or len answer q         answer q append enque    answer   0   if len answer q       answer   int    join  str a  for a in answer q      return answer

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Use itoa to convert to a base 3 string  Drop the last trit and convert back to base 10      Note  itoa is non-standard but actual implementations    don t seem to handle negative when base    10  int div3 int i        char str 42       sprintf str    d   INT MIN      Put minus sign at str 0      if  i gt 0                         Remove sign if positive         str 0             itoa abs i    amp str 1   3         Put ternary absolute value starting at str 1      str strlen  amp str 1        0      Drop last digit     return strtol str  NULL  3      Read back result

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Good  ol bc     num 1337  printf  scale 5   num  x2F3  n    bc 445 66666

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It is easily possible on the Setun computer   To divide an integer by 3  shift right by 1 place   I m not sure whether it s strictly possible to implement a conforming C compiler on such a platform though  We might have to stretch the rules a bit  like interpreting  at least 8 bits  as  capable of holding at least integers from -128 to  127

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Okay I think we all agree that this isn t a real world problem  So just for fun  here s how to do it with Ada and multithreading   with Ada Text IO   procedure Divide By 3 is     protected type Divisor Type is       entry Poke        entry Finish     private       entry Release        entry Stop Emptying        Emptying   Boolean    False     end Divisor Type      protected type Collector Type is       entry Poke        entry Finish     private       Emptying   Boolean    False     end Collector Type      task type Input is    end Input     task type Output is    end Output      protected body Divisor Type is       entry Poke when not Emptying and Stop Emptying Count   0 is       begin          requeue Release        end Poke        entry Release when Release Count  gt   3 or Emptying is          New Output   access Output        begin          if not Emptying then             New Output    new Output              Emptying    True              requeue Stop Emptying           end if        end Release        entry Stop Emptying when Release Count   0 is       begin          Emptying    False        end Stop Emptying        entry Finish when Poke Count   0 and Release Count  lt  3 is       begin          Emptying    True           requeue Stop Emptying        end Finish     end Divisor Type      protected body Collector Type is       entry Poke when Emptying is       begin          null        end Poke        entry Finish when True is       begin          Ada Text IO Put Line  Poke Count Img            Emptying    True        end Finish     end Collector Type      Collector   Collector Type     Divisor   Divisor Type      task body Input is    begin       Divisor Poke     end Input      task body Output is    begin       Collector Poke     end Output      Cur Input   access Input      -- Input value     Number   Integer    18  begin    for I in 1    Number loop       Cur Input    new Input     end loop     Divisor Finish     Collector Finish  end Divide By 3

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Solution using fma   library function  works for any positive number    include  lt stdio h gt   include  lt math h gt   int main         int number   8   Any  ve no      int temp   3  result   0      while temp  lt   number           temp   fma temp  1  3     fma a  b  c  is a library function and returns  a b    c          result   fma result  1  1              printf   n n d divided by 3    d n   number  result       See my another answer

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This will work   smegma  curl http   www wolframalpha com input  i 14 divided by 3 2 gt  dev null   gawk  match  0   link to  input   i   0-9  -      ary    print substr   0  ary 1   start    ary 1   length       4 6666666666666666666666666666666666666666666666666666   Just substitute  14  and  3  with your numbers

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This is a simple function which performs the desired operation  But it requires the   operator  so all you have left to do is to add the values with bit-operators      replaces the   operator int add int x  int y        while  x            int t    x  amp  y   lt  lt  1          y    x          x   t            return y     int divideby3 int num        int sum   0      while  num  gt  3            sum   add num  gt  gt  2  sum           num   add num  gt  gt  2  num  amp  3             if  num    3          sum   add sum  1       return sum       As Jim commented this works  because    n   4   a   b n   3   a    a   b    3  So sum    a  n   a   b  and iterate  When a    0  n  lt  4   sum    floor n   3   i e  1  if n    3  else 0

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int div3 int x      int reminder   abs x     int result   0    while reminder  gt   3           result          reminder--       reminder--       reminder--        return result

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Quite amused none answered with a generic division      For the given integer find the position of MSB    int find msb loc unsigned int n        if  n    0          return 0       int loc   sizeof n     8 - 1      while    n  amp   1  lt  lt  loc            loc--      return loc         Assume both a and b to be positive  return a b    int divide bitwise const unsigned int a  const unsigned int b        int int size   sizeof unsigned int    8      int b msb loc   find msb loc b        int d   0     dividend     int r   0     reminder     int t a   a      int t a msb loc   find msb loc t a       int t b   b  lt  lt   t a msb loc - b msb loc        int i      for i   t a msb loc  i  gt   b msb loc  i--             if  t a  gt  t b                d    d  lt  lt  1    0x1              t a -  t b     Not a bitwise operatiion             t b   t b  gt  gt  1                     else if  t a    t b                d    d  lt  lt  1    0x1              t a   0                    else      t a  lt  t b             d   d  lt  lt  1              t b   t b  gt  gt  1                       r   t a      printf     gt   d  d n   d  r       return d      The bitwise addition has already been given in one of the answers so skipping it

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First that I ve come up with   irb main  101 0 gt  div3   - gt  n   s     0    n to s    s    s       gsub             size     gt    lt Proc 0x0000000205ae90  irb  101  lambda  gt  irb main  102 0 gt  div3 12    gt  4 irb main  103 0 gt  div3 666    gt  222   EDIT  Sorry  I didn t notice the tag C  But you can use the idea about string formatting  I guess

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If you remind yourself standard school method of division and do it in binary  you will discover that in case of 3 you only divide and subtract a limited set of values  from 0 to 5 in this case   These can be treated with a switch statement to get rid of arithmetic operators   static unsigned lamediv3 unsigned n      unsigned result   0  remainder   0  mask   0x80000000        Go through all bits of n from MSB to LSB    for  int i   0  i  lt  32  i    mask  gt  gt   1          result  lt  lt   1         Shift in the next bit of n into remainder      remainder   remainder  lt  lt  1      n  amp  mask           Divide remainder by 3  update result and remainer         If remainder is less than 3  it remains intact      switch  remainder            case 3        result    1        remainder   0        break       case 4        result    1        remainder   1        break       case 5        result    1        remainder   2        break               return result      include  lt cstdio gt   int main          Verify for all possible values of a 32-bit unsigned integer    unsigned i   0     do         unsigned d   lamediv3 i        if  i   3    d              printf  failed for  u   u     u n   i  d  i   3         return 1              while    i    0

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Using Hacker s Delight Magic number calculator   int divideByThree int num      return  fma num  1431655766  0   gt  gt  32       Where fma is a standard library function defined in math h header

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Since it s from Oracle  how about a lookup table of pre calculated answers   -D

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include  lt stdio h gt   include  lt stdlib h gt   int main int argc  char  argv           int num   1234567      int den   3      div t r   div num den      div   is a standard C function      printf   d n   r quot        return 0

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Here s my solution    public static int div by 3 long a        a  lt  lt   30      for int i   2  i  lt   32   i  lt  lt   1            a   add a  a  gt  gt  i             return  int   a  gt  gt  32      public static long add long a  long b        long carry    a  amp  b   lt  lt  1      long sum    a   b       return carry    0   sum   add carry  sum       First  note that   1 3   1 4   1 16   1 64         Now  the rest is simple   a 3   a   1 3   a 3   a    1 4   1 16   1 64        a 3   a 4   a 16   1 64       a 3   a  gt  gt  2   a  gt  gt  4   a  gt  gt  6         Now all we have to do is add together these bit shifted values of a  Oops  We can t add though  so instead  we ll have to write an add function using bit-wise operators  If you re familiar with bit-wise operators  my solution should look fairly simple    but just in-case you aren t  I ll walk through an example at the end   Another thing to note is that first I shift left by 30  This is to make sure that the fractions don t get rounded off   11   6  1011   0110   sum   1011   0110   1101   carry    1011  amp  0110   lt  lt  1   0010  lt  lt  1   0100   Now you recurse   1101   0100   sum   1101   0100   1001   carry    1101  amp  0100   lt  lt  1   0100  lt  lt  1   1000   Again   1001   1000   sum   1001   1000   0001   carry    1001  amp  1000   lt  lt  1   1000  lt  lt  1   10000   One last time   0001   10000 sum   0001   10000   10001   17   carry    0001  amp  10000   lt  lt  1   0  Done    It s simply carry addition that you learned as a child   111  1011  0110 ----- 10001   This implementation failed because we can not add all terms of the equation   a   3   a 4   a 4 2   a 4 3         a 4 i         f a  i    a   1 3   1 4 i f a  i    a 4   a 4 2         a 4 i   Suppose the reslut of div by 3 a    x  then x  lt   floor f a  i    lt  a   3  When a   3k  we get wrong answer

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Didn t cross-check if this answer is already published  If the program need to be extended to floating numbers  the numbers can be multiplied by 10 number of precision needed and then the following code can be again applied      include  lt stdio h gt   int main         int aNumber   500      int gResult   0       int aLoop   0       int i   0      for i   0  i  lt  aNumber  i                  if aLoop    3                       gResult               aLoop   0                      aLoop               printf  Reulst of  d   3    d   aNumber  gResult        return 0

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This one is the classical division algorithm in base 2    include  lt stdio h gt   include  lt stdint h gt   int main       uint32 t mod3 6      0 1 2 0 1 2      uint32 t x   1234567     number to divide  and remainder at the end   uint32 t y   0     result   int bit   31     current bit   printf  X  u   X 3  u n  x x 3      the   3  is for testing    while  bit gt 0          printf  BIT  d  X  u  Y  u n  bit x y          decrement bit     int h   1  while  1    bit    h  if   bit amp h   h  lt  lt   1  else break        uint32 t r   x gt  gt bit      current remainder in 0  5     x    r lt  lt bit              remove R bits from X     if  r  gt   3  y    1 lt  lt bit     new output bit     x    mod3 r  lt  lt bit        new remainder inserted in X       printf  Y  u n  y

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I think the right answer is   Why would I not use a basic operator to do a basic operation

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Using a Linux shell script    include  lt stdio h gt  int main         int number   30      char command 25       snprintf command  25   echo     d  c 3      number  47       system  command        return 0      See my another answer

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To divide a 32-bit number by 3 one can multiply it by 0x55555556 and then take the upper 32 bits of the 64 bit result   Now all that s left to do is to implement multiplication using bit operations and shifts

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Generally  a solution to this would be   log pow exp numerator  pow denominator -1

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note  see Edit 2 below for a better version   This is not as tricky as it sounds  because you said  quot without using the             operators quot   See below  if you want to forbid using the   character all together  unsigned div by unsigned const x  unsigned const by      unsigned floor   0    for  unsigned cmp   0  r   0  cmp  lt   x         for  unsigned i   0  i  lt  by  i          cmp       that s not the   operator      floor   r      r       neither is this        return floor     then just say div by 100 3  to divide 100 by 3   Edit  You can go on and replace the    operator as well  unsigned inc unsigned x      for  unsigned mask   1  mask  mask  lt  lt   1        if  mask  amp  x        x  amp    mask      else       return x  amp  mask        return 0     overflow  note that both x and mask are 0 here      Edit 2  Slightly faster version without using any operator that contains the   -       characters  unsigned add char const zero    unsigned const x  unsigned const y         this exploits that  amp foo bar     foo bar if foo is of type char    return  int  uintptr t   amp    amp zero x   y        unsigned div by unsigned const x  unsigned const by      unsigned floor   0    for  unsigned cmp   0  r   0  cmp  lt   x         cmp   add 0 cmp by       floor   r      r   add 0 r 1         return floor     We use the first argument of the add function because we cannot denote the type of pointers without using the   character  except in function parameter lists  where the syntax type   is identical to type  const  FWIW  you can easily implement a multiplication function using a similar trick to use the 0x55555556 trick proposed by AndreyT  int mul int const x  int const y      return sizeof struct       char const ignore y       x

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The following script generates a C program that solves the problem without using the operators       -        usr bin env python3  print     include  lt stdint h gt   include  lt stdio h gt  const int32 t div by 3 const int32 t input          for i in range -2  31  2  31       print      if input     d  return  d      i  i   3     print r        return 42     impossible   int main         const int32 t number   8      printf   d   3    d n   number  div by 3 number

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First   x 3    x 4     1-1 4    Then figure out how to solve x  1 - y    x  1-1 y      x    1 y     1-y 2      x    1 y     1 y 2     1-y 4              x    1 y     1 y 2     1 y 4           1 y  2 i      1-y  2  i i       x    1 y     1 y 2     1 y 4           1 y  2 i     with y   1 4   int div3 int x        x  lt  lt   6        need more precise     x    x gt  gt 2      x   x    1  1 2  2      x    x gt  gt 4      x   x    1  1 2  4      x    x gt  gt 8      x   x    1  1 2  8      x    x gt  gt 16     x   x    1  1 2  16      return  x 1  gt  gt 8     as  1- 1 2  32  very near 1                          we plus 1 instead of div  1- 1 2  32      Although it uses    but somebody already implements add by bitwise op

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I would use this code to divide all positive  non float numbers  Basically you want to align the divisor bits to the left to match the dividend bits  For each segment of the dividend  size of divisor  you want to check to make sure if the the segment of dividend is greater than the divisor then you want to Shift Left and then OR in the first registrar  This concept was originally created in 2004  standford I believe   Here is a C version which uses that concept  Note   I modified it a bit   int divide int a  int b        int c   0  r   32  i   32  p   a   1      unsigned long int d   0x80000000       while   b  amp  d     0                d  gt  gt   1          r--             while  p  gt  a                c  lt  lt   1          p    b  gt  gt  i--   amp    1  lt  lt  r  - 1           if  p  gt   a              c    1            return c    p is remainder  for modulus      Example Usage   int n   divide  3  6     outputs 2

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Where InputValue is the number to divide by 3  SELECT AVG NUM     FROM  SELECT InputValue NUM from sys dual          UNION ALL SELECT 0 from sys dual          UNION ALL SELECT 0 from sys dual  divby3

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bin ruby  def div by 3 i    i div 3          always return int http   www ruby-doc org core-1 9 3 Numeric html method-i-div end

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include  lt stdio h gt   typedef struct   char a b c    Triple   unsigned long div3 Triple  v  char  r      if   long v  lt   2        return  unsigned long r    return div3  amp v -1    amp r 1       int main       unsigned long v   21     int r   div3  Triple  v  0      printf   ld   3    d n   v  r      return 0

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log pow exp number  0 33333333333333333333       -

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It seems no one mentioned the division criterion for 3 represented in binary - sum of even digits should equal the sum of odd digits  similar to criterion of 11 in decimal   There are solutions using this trick under Check if a number is divisible by 3   I suppose that this is the possible duplicate that Michael Burr s edit mentioned

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It s really quite easy   if  number    0  return 0  if  number    1  return 0  if  number    2  return 0  if  number    3  return 1  if  number    4  return 1  if  number    5  return 1  if  number    6  return 2     I have of course omitted some of the program for the sake of brevity   If the programmer gets tired of typing this all out  I m sure that he or she could write a separate program to generate it for him  I happen to be aware of a certain operator     that would simplify his job immensely

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You can use  platform dependent  inline assembly  e g   for x86   also works for negative numbers    include  lt stdio h gt   int main       int dividend   -42  divisor   5  quotient  remainder       asm      cdq  idivl   ebx                 a   quotient     d   remainder               a    dividend    b    divisor                    printf   i    i    i  remainder   i n   dividend  divisor  quotient  remainder     return 0

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Well you could think of using a graph tree like structure to solve the problem  Basically generate as many vertices as the number that is to be divided by 3  Then keep pairing each un-paired vertex with two other vertices   Rough pseudocode   function divide int num      while num  0          Add a new vertice to vertiexList          num--     quotient   0     for each in vertexList lets call this vertex A          if vertexList not empty             Add an edge between A and another vertex say B          else             your Remainder is 1 and Quotient is quotient         if vertexList not empty             Add an edge between A and another vertex say C          else             your remainder is 2 and Quotient is quotient         quotient           remove A  B  C from vertexList     Remainder is 0 and Quotient is quotient   This can obviously be optimized and the complexity depends on how big you number is  but it shoud work providing you can do    and --  It s as good as counting only cooler

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3 in base 2 is 11   So just do long division  like in middle school  in base 2 by 11   It is even easier in base 2 than base 10   For each bit position starting with most significant   Decide if prefix is less than 11   If it is output 0   If it is not output 1  and then substitute prefix bits for appropriate change  There are only three cases    11xxx - gt     xxx     ie 3 - 3   0  100xxx - gt    1xxx     ie 4 - 3   1  101xxx - gt   10xxx     ie 5 - 3   2    All other prefixes are unreachable   Repeat until lowest bit position and you re done

[c] Divide a number by 3 without using *, /, +, -, % operators

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