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 |
Answer : 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 2n-1 which is greater than or equal to that is 19,342,813,113,834,066,795,298,815. That corresponds to 284-1. 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. ARUMUGAM said: (4:18pm on Wednesday 15th 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
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Option: A Explanation : Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. |
