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Solution

For 8-bit signed integer (two’s complement):

Range = $-2^{7}$ to $2^{7}-1$

Least value = $-128$

For 8-bit unsigned integer:

Range = $0$ to $2^{8}-1$

Least value = $0$

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Solution

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Solution

A black and white image needs $1$ bit per pixel.
Size of one $1$-megapixel black and white image $=1\times 10^6$ bits.
Size of $8$ such images $=8\times 10^6$ bits $=1\ MB$.
An $8$-color image needs $\log_2 8=3$ bits per pixel.
Size of one $8$-megapixel $8$-color image $=8\times 10^6\times 3=24\times 10^6$ bits $=3\ MB$.
Total space $=1\ MB+3\ MB=4\ MB$.

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Solution

In base $y$, digits must be less than base.

Number $(45)_y$ has digit 5.

Thus

$y>5$

Possible bases:

$6,7,8,\dots$

Given options:

6, 7, 8

Thus

(B), (C), (D)