Digital-Logic-Design
October 16, 2023PUERTA 2008-IT
October 16, 2023Digital-Logic-Design
| 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 2n -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 2n -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 2n -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 2n -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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