A decimal number has 25 digits. The number of bits needed for its equivalent binary representation is, approximately
A.  50 
B.  55 
C.  85 
D.  80 
Option: C Explanation : The largest decimal number with 25 digits is 9,999,999,999,999,999,999,999,999. The smallest decimal number in the form 2^{n}1 which is greater than or equal to that is 19,342,813,113,834,066,795,298,815. That corresponds to 2^{84}1. So, the minimum number of binary bits required to represent the decimal number 25 nines in a row is 84. This is 84 ones in a row. If you want to support negative as well as positive numbers, you will need 85. Click on Discuss to view users comments. HEMANT SINGH said: (2:17am on Friday 10th May 2013)
how to get it
ARUMUGAM said: (2:18am on Thursday 16th May 2013)
Generalised Expalanation:Let k bits are required to represent d digit decimal number.(ie) Let 2 power k = 10 power dApply base 2 logarithm on both sideslog(2 power k) base 2 = log (10 power d) base 2k log2 base 2 = d log 10 base 2As log10 base 2=3.32192 and log2 base2 =1,k=3.32192dHere d=25,k=3.32192*25k=83

A.  10000 
B.  01011 
C.  01100 
D.  01101 
Option: B Explanation : Click on Discuss to view users comments. 
The decimal number 80 can be represented in BCD code as
A.  1000 0001 
B.  0101 0000 
C.  0010 0001 
D.  1000 0000 
Option: D Explanation : Click on Discuss to view users comments. 
How many fractional digits 248 will have?
A.  24 
B.  47 
C.  49 
D.  48 
Option: D Explanation : Click on Discuss to view users comments. 
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