in Digital Logic edited by
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The largest positive number in 2's complement format represented with 8-bits is:

  1. $(\mathrm{FF})_{16}$
  2. $(128)_{10}$
  3. $(777)_8$
  4. $(01111111)_2$
in Digital Logic edited by
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5 Comments

there we no options provided to select it was given as numerical type....

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@Lakshman Bhaiya 
please change the question type to MSQ from NAT..

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@shadymeee can you please confirm the below scenario?
if the question is about "
The largest positive number in 2's complement representation, represented with 8-bits is:" then A will be the answer?

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for the number to bo posotive in n bits its signbit must be 0, hence we know that largest positive number in 2s complement for n bit is 2^(n-1)-1,

so 2^(8-1)-1.
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oh! got confused in 2's complement and 2's complement representation as in this question.
https://gateoverflow.in/367765/topic-number-systems-digital-logic
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1 Answer

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0 votes

A cool way to solve this problem is,

we have a 8 bits binary number which is in 2's complement form.

Now the 8th bit that is the MSB bit, represents the sign of the number. 

So, 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1

now with 8 bits we have the total range as 

10000000 = -128 (smallest number) to 01111111 = 127 (largest number)

So we have range as -128, -127, -126, -125,... 0, 1, 2, 3, 4, 5,... 127

as the last option is clearly the last largest number in our range therefore it's the answer.

Option D.

Answer:

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