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We have 32 bit instruction

We need  7 bits for specifying an particular register address ($\left \lceil log_{2}100 \right \rceil$)

We need 20 bits to address a particular word in memory $\left \lceil log_{2}1 M \right \rceil$

out of 32 if we subtract k and L address  would get  0 address instruction

$2^{32}-k*2^{7}*2^{20}-L*2^{7}$

$((2^{5}-k)2^{20}-L)2^{7}$

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