9 votes 9 votes The minimum number of $\text{NAND}$ gates required to implement the Boolean function $A + A\overline{B} + A\overline{B}C$ is equal to $0$ (Zero) $1$ $4$ $7$ Digital Logic digital-logic min-no-gates isro2016 + – neha singh asked Mar 25, 2016 edited Dec 7, 2022 by Lakshman Bhaiya neha singh 7.1k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply KBhuvana commented Apr 28, 2017 reply Follow Share Is it required to simplify expression before constructing it? Because without simplifying the expression we can have construction with 7 NAND gates... so getting confused. Please clarify 0 votes 0 votes Samujjal Das commented Apr 28, 2017 reply Follow Share Yes it is always required to simplify the expression. We are not interested in the expression actually. We are interested in the output. So whatever is the output of the longer expression is also the output of the simplified one. So why not simplify the expression and realise it!! The number of devices required will be less and we are always interested in optimisation. :-) 0 votes 0 votes Please log in or register to add a comment.
Best answer 24 votes 24 votes ZERO Because After simplifying You would get This A(1+B'+B'C) which is equal To A So No need For any NAND gate bad_engineer answered Mar 25, 2016 selected Jul 4, 2016 by srestha bad_engineer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments cse23 commented Mar 30, 2016 reply Follow Share I think 2 NAND gates as A itself can be represented by 2 NAND gates. 1 votes 1 votes Registered user 7 commented Jul 4, 2016 reply Follow Share is it correct 0 votes 0 votes Warrior commented May 1, 2017 reply Follow Share yes it is Correct. 0 votes 0 votes Please log in or register to add a comment.