Question

Suppose the largest n bit number requires 'd' digits in decimal representation. Which of the following relations between 'n' and 'd' is approximately correct?

• A. \( d = 2^n \)
• B. \( n = 2^d \)
• C. \( d < n \log_{10} 2 \)
• D. \( d > n \log_{10} 2 \)

Hint:

To relate the number of bits (\(n\)) to the number of decimal digits (\(d\)), equate the total number of representable values: \( 2^n \approx 10^d \). Taking \( \log_{10} \) of both sides gives \( n \log_{10} 2 \approx d \). Since \( d \) must be an integer, \( d = \lceil n \log_{10} 2 \rceil \), which generally means \( d \) is slightly greater than \( n \log_{10} 2 \).

Answer 1

The Correct Option is D

The largest number that can be represented with \( n \) bits is \( N = 2^n - 1 \).
The number of decimal digits, \( d \), required to represent an integer \( N \) is given by the formula \( d = \lfloor \log_{10} N \rfloor + 1 \).
For large values of \( n \), \( 2^n - 1 \) is very close to \( 2^n \). So we can approximate \( N \approx 2^n \).
Substituting this into the formula for \( d \): \[ d \approx \lfloor \log_{10} (2^n) \rfloor + 1 \]
Using the logarithm property \( \log(a^b) = b \log(a) \): \[ d \approx \lfloor n \log_{10} 2 \rfloor + 1 \]
Since \( \log_{10} 2 \approx 0.3010 \), the relation is approximately \( d \approx \lfloor 0.301n \rfloor + 1 \).
This means \( d \) will be an integer slightly greater than \( n \log_{10} 2 \).
For example, if n=10, \( n \log_{10} 2 \approx 3.01 \). The number of digits required for \( 2^{10}-1=1023 \) is 4. Clearly, \( 4 > 3.01 \). If n=16, \( n \log_{10} 2 \approx 4.81 \). The number of digits required for \( 2^{16}-1=65535 \) is 5. Clearly, \( 5 > 4.81 \).
In general, \( d = \lceil n \log_{10} 2 \rceil \) (ceiling function), which implies \( d \geq n \log_{10} 2 \). The strict inequality \( d > n \log_{10} 2 \) holds unless \( n \log_{10} 2 \) is an integer, which is rare. So, this is the best approximation.