in Digital Logic edited by
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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

  1. $0$ (Zero)
  2. $1$
  3. $4$
  4. $7$
in Digital Logic edited by
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2 Comments

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
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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. :-)
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1 Answer

24 votes
24 votes
Best answer

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

 

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4 Comments

I think 2 NAND gates as A itself can be represented by 2 NAND gates.
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is it correct
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yes it is Correct.
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