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P is a 16-bit signed integer. The 2’s complement representation of P is $(E77B)_{16}$. The 2's complement representation of 8×P is?
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Multiplication can be directly carried in 2's complement form. E77B = 1110 0111 0111 1011 can be left shifted 3 times to

give 8P =  0011 1011 1101 1000    = 3BD8.

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But now, the MSB is 0 which means it is a positive number.

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