Ace Test Series: Digital Logic - ROM

Question

I am getting (C). But answer given is (B). Where I have gone wrong?

Comments

Sorry sir :( I just wanted to clear my confusion

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0

1

0100

1001

10000

.....

11010001 (225)

Totally 16 selections needed and we select each using 4 bits. Each selection can be up to 8 bits. So, answer is B. 

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Ohkay. Got it! Actually I was confused between multiplier and squarer :p

Answer 1

A  $2^{k}\times n$ ROM required $k\times 2^{k}$ decoder and $n$ OR gates

output square of $4$ bit numbers, So, input 4 bit decoder and with this 4 bit we can represent $(1111)_{2}=15$ in decimal

Now, square of $15^{2}=225$ which can be represented by $8$ bits

So, size of ROM will be $2^{4}\times8$

$4$ address lines $8$ data lines