(A . A)' = A'

Now, we can get A' by using 1 NAND gate.

Similarly, we can get B' by using 1 NAND gate.

Now, we have to perform 1 NAND operation for which we need 1 NAND gate. We get (A'B')' as the output.

Now, we have to take complement of this output for which we need 1 NAND gate. Now, the output is ((A'B')')'.

((A'B')')' = (A+B)'

So, totally, we need 4 NAND gates in order to build a NOR gate.

Ans.: **4**