For a 4-bit multiplier there are $2^4 \cdot 2^4 = 2^8$ combinations.
The output of 4-bit multiplication is 8 bits, so the amount of ROM needed is $2^8 \cdot 8 = 2048$ bits.
Why is that? Why does the ROM need all the combinations embedded into it?
What will be the case with RAM?
Question Source : http://cs.stackexchange.com/questions/9471
As shown in the figure bellow (general architecture of ROM):
to address 8-bit numbers correctly your ROM size should be: $2^8 \times 8$ bits.
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