Arjun Suresh sir- Can you tell the book where this topic is covered. It will be great if you can provide some link.

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+29 votes

The amount of ROM needed to implement a $4-bit$ multiplier is

- $64$ bits
- $128$ bits
- $1$ Kbits
- $2$ Kbits

+37 votes

Best answer

A ROM cannot be written. So, to implement a $4$-bit multiplier we must store all the possible combinations of $2^4 \times 2^4$ inputs and their corresponding $8$ output bits giving a total of $ 2^4 \times 2^4 \times 8$ bits $= 2048$ bits. So, (**D**) is the answer.

PS: We are not storing the input bits explicitly -- those are considered in order while accessing the output $8$ bits. In this way, by storing all the possible outputs in order we can avoid storing the input combinations.

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Arjun Suresh sir- Can you tell the book where this topic is covered. It will be great if you can provide some link.

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I guess Morris Mano book covers this. I don't have the book now to verify. This ppt also gives a very good detail:

http://www.cs.uiuc.edu/class/fa05/cs231/lectures/review(11-4).ppt

http://www.cs.uiuc.edu/class/fa05/cs231/lectures/review(11-4).ppt

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@Arjun Suresh Sir, I have a doubt here... With 4 bits, 16 numbers (multiplicands) are possibe. Each of this number takes 4 bits in memory.. Similarly each multiplier will also take 4 bits then why dont we multiply the result u found with 4*4... 2^4 is just number of multiplicands.. It is not number of bits used to store those many multiplicands... Please clarify

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