DigitalLogicDesign
October 16, 2023PUERTA 2008IT
October 16, 2023DigitalLogicDesign
Question 430

A decimal has 25 digits. the number of bits needed for its equivalent binary representation is approximately
50


74


40


60


None of these

Question 430 Explanation:
Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 10^{ 3} 1 which is 999.
Then, Decimal number has 25 digits, so maximum number is 10^{ 25} 1
Similarly, in the binary representation with “n” bits the maximum number is 2^{n} 1
So we can write 10^{ 25} –1 = 2^{ n} – 1 —>10 ^{25} = 2^{ n}
After taking log ^{2} on both sides
log _{2} 2^{ n} =log_{ 2} 10^{ 25}
n log _{2} 2=25 log_{ 2} 10
n=25*(3.322) [ log _{2} 2=1 & log_{ 2} 10 =3.322]
n=83
Then, Decimal number has 25 digits, so maximum number is 10^{ 25} 1
Similarly, in the binary representation with “n” bits the maximum number is 2^{n} 1
So we can write 10^{ 25} –1 = 2^{ n} – 1 —>10 ^{25} = 2^{ n}
After taking log ^{2} on both sides
log _{2} 2^{ n} =log_{ 2} 10^{ 25}
n log _{2} 2=25 log_{ 2} 10
n=25*(3.322) [ log _{2} 2=1 & log_{ 2} 10 =3.322]
n=83
Correct Answer: E
Question 430 Explanation:
Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 10^{ 3} 1 which is 999.
Then, Decimal number has 25 digits, so maximum number is 10^{ 25} 1
Similarly, in the binary representation with “n” bits the maximum number is 2^{n} 1
So we can write 10^{ 25} –1 = 2^{ n} – 1 —>10 ^{25} = 2^{ n}
After taking log ^{2} on both sides
log _{2} 2^{ n} =log_{ 2} 10^{ 25}
n log _{2} 2=25 log_{ 2} 10
n=25*(3.322) [ log _{2} 2=1 & log_{ 2} 10 =3.322]
n=83
Then, Decimal number has 25 digits, so maximum number is 10^{ 25} 1
Similarly, in the binary representation with “n” bits the maximum number is 2^{n} 1
So we can write 10^{ 25} –1 = 2^{ n} – 1 —>10 ^{25} = 2^{ n}
After taking log ^{2} on both sides
log _{2} 2^{ n} =log_{ 2} 10^{ 25}
n log _{2} 2=25 log_{ 2} 10
n=25*(3.322) [ log _{2} 2=1 & log_{ 2} 10 =3.322]
n=83
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