Discussion
Minimum number of bits required to store any 3 digit decimal number is equal to
Answer: D
The greatest three digit decimal number is 999, so the number of bits needed is the smallest n with \(2^n\) greater than 999:
\[ 2^{9} = 512 < 999 \]
\[ 2^{10} = 1024 > 999 \]
So 10 bits are needed, since 9 bits can store only the numbers up to 511 and 10 bits go up to 1023. The options 3, 5 and 8 are too few, since 8 bits can hold only numbers up to 255, and 3 would be the number of decimal digits and not of bits. The answer is 10.
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