问题代码：

; Copy all the values which are divisible by four to a starting address $1600.
; The indivisible by four is pushed onto the stack, and use the effective address
; to pull them out after finishing the loop. The starting address is $1650.
; Array pointer starts from the beginning, and using DBNE to decrement the loop

N       equ     12
        org     $1600
        fill    0,128           ; each row in the memory displays 16 bytes
cnt     rmb     1               ; indivisible counter
        org     $1ff0
        fill    0,16

        org     $2000           ; start RAM address
        lds     #$2000          ; initialize the stack pointer
        ldy     #$1600          ; load the starting address for divisible by 4
        ldx     #array          ; load the starting address of array to Reg. X
        ldaa    #N              ; use B as the loop counter, B = N
        clr     cnt             ; clear cnt counter
loop    brclr   0,x,$03,yes     ; if bits 1 & 0 are 0s, then branch to yes
        ldab    1,x+
        inc     cnt             ; increment cnt counter for indivisible
        pshb                    ; push onto the stack for the indivisible
        bra     chkend
yes     movb    1,x+,1,y+       ; copy the divisible by 4 value to a new location
chkend  dbne    a,loop          ; decrement B by 1, if B NE 0, branch to loop

; use the effective address to load/pull the elements from the stack

        ldy     #$1650
        ldx     #$1fff          ; first available address for the stack
        ldaa    cnt
loop2   movb    1,x-,1,y+       ; stack is first in last out
        dbne    a,loop2
        swi
array   db      3,4,9,12,19,20,24,31,48,53,64,72
        end

我正在学习微处理器，我的教授给了我们这个代码来修改，但我不确定如何修改。这是我们第一次用汇编语言“编程”。
他还来帮我们这个忙：

EEEN 3449
Lab 4c:
• idiv => Accumulator D / Register X, the remainder refreshes to D
• Register X has the constant 3.
• Register Y is utilized as the array pointer.
• Although there is nothing stored in Accumulator A, A:B => D
• You can use Accumulator B as the loop counter, and store the
starting address for divisible by 3.
• In the beginning of the loop, you have to store loop counter value in
another temporary address, like $15FE.
• clr A, and ldab 0,y will generate [D] = [0 : 1st array element]
• Carry out the idiv and compare Accumulator D with 0.
• If Accumulator D is non-zero, push the accumulator B onto the stack.
• If Accumulator D is zero, reload the updated address for divisible by 3.
• After sorting out the array element where to store them, reload the
loop counter from $15FE, and increment the array pointer.
• Then use dbne b,loop to execute the next iteration.

由于不知道汇编，我不知道如何使用它来正确地更改代码。
显示的代码是拉的数字是可被4整除，并把他们放在内存中的$1600，我想。我需要改变这个代码，使它拉的数字是可被3整除。没有线索从哪里开始，并开始测试，因为我不知道一个完整的汇编很多。
如果你能帮助我理解正在发生的事情，任何帮助都是很好的。我确实想学习，但是我似乎找不到很多关于汇编的东西。
我试过自己测试东西，但我总是得到错误，不知道他们在哪里，也不知道如何修复他们。