## Discussion Forum

Que. | What is the minimum number of gates required to implement the Boolean function (AB+C)if we have to use only 2-input NOR gates? |

a. | 2 |

b. | 3 |

c. | 4 |

d. | 5 |

Answer:3 |

Ramesh :(November 18, 2021)
To me, the answer is 4. Applying De Morgans theorem two times will lead to y'' = ((A'+B')+C')'
This requires 4 NOR gate, 1 for A', another for B', another for C' and the last for ((A'+B')+C')' review this answer please |

AB+C = (A+C)(B+C) = ((A+C)’ + (B+C)’)’. So 3 NOR gates are required. |

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