For disk size of $768 \hspace{0.1cm}MB$ and block size of $384 \hspace{0.1cm} bytes$ what would be the bitmap size in bytes(as a power of $2$)
iI represented $768 \hspace{0.1cm} MB$ as $3*2^{28}$ and $384$ as $3*2^7$ and then $\dfrac{3*2^{28}}{3*2^7}= 2^{21}$ but the correct answer given was not $2^{21}$ but $2^{18}$ because he divides $2^{21}/2^3$ and the explanation is $2^3$ is how many bits are in a byte, but why do this division isnt representing $768 \hspace{0.1cm} MB$ as $768*2^{20}$ or $3*2^{28}$ already in bytes, after all MB has $2^{20} \hspace{0.1cm} bytes$ so the answer $2^{21}$ should already be in bytes, right?