4. Minterm m31 must be represented by at least how many literals?

  1. 6
  2. 31
  3. 3
  4. 4
  5. 5

5 is CORRECT. It takes at least 5 literals to represent m31 because 5 literals are required to form 2^5 = 32 minterms. Also observe that the minterm code for m31 is 11111; since (11111)2 =(31)10. Thus 5 literals (variables or their complements) are required to support this 5 bit code.

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