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What is the minimum size of ROM required to store the complete truth table of an $8-bit \times 8-bit$ multiplier?

  1. $32 K \times 16$ bits
  2. $64 K \times 16$ bits
  3. $16 K \times 32$ bits
  4. $64 K \times 32$ bits
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repeated question in GATE 2019 or i think gate 2020
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Best answer
Answer - B.

Multiplying $2\ 8$ bit digits will give result in maximum $16$ bits

Total number of multiplications possible $= 2^8 \times 2^8$

Hence, space required $= 64K \times 16$ bits
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They have also mentioned the MINIMUM.

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so what minimum

answer will be what sachin has said.
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My dear friend  @ , i said minimum because even 64K x 32 bit ROM can also store the given information but it will not be the minimum size ROM to perform this job. So please don’t jump to your conclusions straight away.

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3 votes
3 votes
I think the answer should be 64K x 32 bits. Because we have an output for each of 2^8 x 2^8 combinations of inputs. But in question, they have asked to store the total truth table which includes inputs as well to tell which output corresponds to which input

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Yeah exactly what I was thinking.
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