Hi, All

This question has been asked me in last 3 exams and i still does not have solution. I need to multiply 3 any bunary numbers.
They does not want to do partial product method. They have program writen so we have to do that way. Program is simple, it will keep adding multipliplier times multiplicand.
i.e. 3x4
= 3+3+3+3
= 12
So i need 8 bit adder and 8 bit register. Register input will be adder output and register output will feedback in to the adder input.. I just need logic that will count down(multiplicand) and stop at 0000, so i can use that output to give input to Register to load pin. so It will stop adding and give output as product.

If anyone have any other ideas please let me know. I can use only d flip flop and basic gates. Sorry to be so long.

 

You can multiply any binary number by 3 by shifting left 1 place (x 2) and adding the original number.

So create an n-bit full adder - which can be done with basic gates.
Into its A inputs connect your binary input.
Into its B inputs connect your binary input shifted 1 place left. This could be done simply by wiring but you could construct a parallel-load shift register out of your D-type flip-flops.
The adder output will the input times 3.

Is this what you mean?

 

No, I want multiplication any 3 digit binary number. So it can be any numbers. Like 8x4 or any.. And Multiplication has to be done by adding the number not by partial product method.
So lets say 8 multiplier and 4 is multiplicand so

8+8+8+8=32

Well anyway I solved the question and working good..

Thanks for reply.