MADEEASY TESTSERIES DLD

Question

The representation of the value of 20 bit signed integer in 2’s complement form is P = (A72E5)₁₆ . Which of the following represents 16 × P in 1’s complement representation?

A)(72E4F)₂₀

B)(72E50)₂₀

C)(72E4E)₂₀

D)NONE OF above

ON seeing options i marked option d becz none of these have base 16....but actual answer given is option A ...no reson given but what i feel is that base greater than 15 should be there????no matter 16 or 20???

Comments

Shaik Masthan but 72E50 is 470608 here you can see that MSB is 1 so that's a negative number so we need to subtract 1 from it giving option a

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In 72E50, MSB bit is 1??

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I'm sorry. you're correct.

Answer 1

Option a is right.

• In multiplication of 2's complement .

• We perfome logical left shift operation.

• I.e remove MSB bit and add Zero in LSB .

• In this question we need to multiply with 16 . Then remove 4  MSB bit and add  4 zero in LSB bits . Teen we get 16 *p . 

• 16*p is 2's complementation . Then we subtract one we get 1's complement.

Comments

@ bhishekmehta4u and @ Shaik Masthan what is the correct procedure in case of loss of bits is not allowed due to msb contains 1 or number of shifts causes overflow 

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@ eyeamgj , i am not getting your question

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@ Shaik Masthan  i m saying suppose take a question whose first four msb contains all zeros so as on doing left shifting no problem will there??  and what if msb are like this question so what then ???