In the contest! :)

 movwf  reg0      ;reg0=3w/8
 bcf    status,c
 rrf    reg0,f
 addwf  reg0,f
 rrf    reg0,f
 bcf    status,c
 rrf    reg0,f
 movf   reg0,w
 movwf  reg1,f    ;iteratively refine twice
 bcf    status,c  ;w/3=3w/8-(2(w'/3))/16
 rlf    reg1,f
 swapf  reg1,w
 andlw  0x0f
 subwf  reg0,w
 movwf  reg1,f
 bcf    status,c
 rlf    reg1,f
 swapf  reg1,w
 andlw  0x0f
 subwf  reg0,w

Enter with a byte in w, exit with the byte divided by 3 in w.
20 words program, 20 cycles, 2 register bytes.

If you want to save 3 words make the iteration a subroutine,
this will cost you 8 cycles in execution time.

Anyone came up with a better code? (I quit this subject...) ;)

-- Lauri

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