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A $\text{ROM}$ is used to store the table for Addition of $2$ number of $16$ bit unsigned integers. The size of the $\text{ROM}$ required for the above application in Giga Bytes is _______________.
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Ohh thanks @shubham

I have calculated in bits :(

68Gb = 8.5GB
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Just small additional info: For all such kinds of questions, Size of ROM = 

Size of 1 result * Number of all possible results

Here, same question but with multiplication https://gateoverflow.in/2725/gate1996-1-21

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Best answer
 

 total entry=216*216=232

 

 size of each entry is....Max(n,m) bits

therefor ...total size =232*16/8 =8Gbytes 

considering carry:- 232 *17/8 =8.5 Gbytes

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