Since this problem involves small numbers (particularly with a small loop count of 100), it's possible to ease the modulo operation setup by simply working with 16-bit and 8-bit registers:$$\dfrac{\text{[AX] (16-bit register)}}{\text{[other 8-bit register]}} = \text{[AH] (remainder)}$$My main concern is with the layout.  Every basic high-level implementation I've seen has a check and print together for each case.  I've found it easier to do the same thing here, but I'm not sure if that would also be too readable in assembly.I'm also aware that it's good to minimize register-moving.  Unfortunately, I still do that with every case since I'm incrementing with one register (CX) and using another (AX) for the dividend.  I can stick to AX for both, but that may involve keeping a copy of the current counter value, which may just make the code a bit more complicated.  I suppose it's not much of a problem here anyway.Macros used:nwln -  prints a newlinePutStr - prints a defined stringPutInt - prints a 16-bit integer valueIt's not necessary to address the macros; they do work properly.%include macros.s.DATAfizz_lbl:       DB    Fizz, 0buzz_lbl:       DB    Buzz, 0fizzbuzz_lbl:   DB    FizzBuzz, 0.CODE.STARTUPxor   CX, CX ; countermain_loop:   inc   CX   cmp   CX, 100   jg    done   fizzbuzz_check:      mov   AX, CX ; dividend = counter      mov   BH, 15 ; divisor      div   BH     ; (counter / 15)      cmp   AH, 0            ; counter divisible by 15?      je    print_fizzbuzz   ; if so, proceed with printing      jmp   fizz_check       ; if not, try checking for fizz      print_fizzbuzz:          PutStr   fizzbuzz_lbl          nwln          jmp      main_loop   fizz_check:      mov   AX, CX ; dividend = counter      mov   BH, 3  ; divisor      div   BH     ; (counter / 3)      cmp   AH, 0        ; counter divisible by 3?      je    print_fizz   ; if so, proceed with printing      jmp   buzz_check   ; if not, try checking for buzz      print_fizz:          PutStr   fizz_lbl          nwln          jmp      main_loop   buzz_check:      mov   AX, CX ; dividend = counter      mov   BH, 5  ; divisor      div   BH     ; (counter / 5)      cmp   AH, 0        ; counter divisible by 5?      je    print_buzz   ; if so, proceed with printing      jmp   print_other  ; if not, then can only display number      print_buzz:          PutStr   buzz_lbl          nwln          jmp      main_loop   print_other:      PutInt   CX      nwln      jmp   main_loopdone:   .EXIT
16-bit FizzBuzz in x86 NASM assembly
assembly;fizzbuzz
Since we're doing this in assembly language, it makes sense to do it much more efficiently than is typically done in high level languages.  Otherwise, why bother with assembly language? So with that said, there are ways that this can be made much, much more efficient.Avoid divisionThe div instruction in x86 is one of the slower instructions possible.  Since we already know that we're looking for numbers divisible by 3, 5 or both, what would make far more sense is to simple keep countdown counters for both.  Your initialization currently says:    xor cx, cxIt could be easily expanded to say:    xor cx, cx    mov bx,0503h  ; set bh = 5 counter, bl = 3 counterThen instead of dividing, simply decrement:    inc cx    cmp cx, 100    jg done    dec bh    dec bl    cmp bx, 0    je print_fizzbuzz    cmp bl, 0    je print_fizz    cmp bh, 0    je print_buzzprint_other:Naturally the various print_... routines would have to reset bh, bl or both as well as printing.Improve formattingGenerally speaking, assembly language code is not indented in the way you have your code indented.  It's much more linear, with the only indentation for assembly language statements or directives.Consider better I/OYour output routines are not shown, but it's likely that it would be more efficient to keep the numeric output in string form, incrementing each ASCII digit and emitting the string, rather than repeatedly converting from binary register contents to a string value.