EMU8086 dividing 32 bit number by a 16 bit number gives unexpected 0 remainder

Question

I was trying to divide (Unsigned) 8A32F4D5 by C9A5 using emu8086 tool. I expected the quotient to be AF73H and the remainder be 94B6H. After writing the following code, I was getting correct quotient but the remainder became 0000h. Am I missing something?

.MODEL SMALL
.STACK 100H
.DATA 
.CODE 

MAIN PROC 
; initialize DS
MOV AX,@DATA 
MOV DS,AX 
; enter your code here
MOV DX, 8A32H
MOV AX, 0F4D5H 
MOV BX, 0C9A5H

DIV BX

;exit to DOS 
               
MOV AX,4C00H
INT 21H 

MAIN ENDP
    END MAIN 

The output in EMU8086:

Answer 1

This looks like a bug in EMU8086. There is no division by zero nor is there an overflow with this unsigned division (DIV). You are correct that 0x8A32F4D5 divided by 0xC9A5 has a remainder of 0x94B6. To verify this I ran this code with Turbo Debugger in DOSBOX and got the expected results:

Had this been signed division using the IDIV instruction it would produce a division by zero exception because of division overflow.