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A ROM is used to store the table for multiplication of two $8$-bit unsigned integers. The size of ROM required is

  1. $256 \times 16$
  2. $64 K \times 8$
  3. $4 K \times 16$
  4. $64 K \times 16$
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When we multiply two $8$ bit numbers result can go up to $16$ bits. So, we need $16$ bits for each of the multiplication result. Number of results possible $= 2^8 \times 2^8 = 2^{16} = 64 \ K$ as we need to store all possible results of multiplying two $8$ bit numbers. So, $64 \ K \times 16 $is the answer.

Correct Answer: $D$
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  • Two 8-bit unsigned integers can each have $2^8 = 256$ possible values.
  • The total number of possible input combinations is $256 *256 = 65,536$.
  • The product of two $8-bit$ numbers can be up to $16 bits$ wide ($8 bits + 8 bits = 16 bits$).
  • To access $65,536$ unique combinations, we need $16$ address lines ($2^16 = 65,536$).
  • To store $16-bit$ values, we need 16 data lines.
  • $16$ address lines to specify the input combination (65,536 possible combinations).
  • $16$ data lines to output the 16-bit product value.
  • 64K x 16, where $64K$ represents the number of addressable locations (65,536) and $16$ represents the width of each location (in bits)
Answer:

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