You can make a divider with multiple subtractions. The method is simple: first you try to subtract dividend by divisor x nth power of 2. In here n is the bit number (4-bit, 8-bit or else) then if you can subtract the first bit of solution is 1 else 0 (1xxx or 0xxx for 4-bit for example). Than if it's 1 you alsa make subtraction else don't do it. At the end decrease n by one by one and do this every time. For example:

4-Bit Divison (15 divided by 3):

• Is 15 can be subtract by 3 x 8: No. Then our answer is 0xxx,
• Is 15 can be subtract by 3 x 4: Yes. Then our answer is 01xx and now 15 goes to 3.
• Is 3 can be subtract by 3 x 2: No. Then it's 010x
• Is 3 can be subtract by 3 x 1: Yes. Then our final answer is 0101 => Decimal 5 and it's true.