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The decimal number has $64$ digits. The number of bits needed for its equivalent binary representation is?

  1. $200$
  2. $213$
  3. $246$
  4. $277$
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4 Answers

Best answer
31 votes
31 votes
The concept is "maximum value of 10 digit number represented by 64 digits should be equal to x digits number in base 2."

For maximum value with 64 digits, each digit must be 9. This is given by $10^{64} - 1$. Similarly, maximum value of $x$ digits in binary is $2^x - 1$.

So $10^{64}-1 = 2^x-1$

$10^{64} = 2^x$

$x = 64 . \log_2 10$

$x = 64 \times 3.32 \approx 213.$
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5 votes
5 votes

Suppose n bit binary number requires d digit in decimal representation.

Then relationship can be shown as

d>nlog1o 2

Here d=64

64=n*.3010

So n=64/.3010

     n=213 (apx)

Reference :Number of Digit in Binary

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3 votes
3 votes

The concept is the maximum number of 10 digit number by 10 digit should be equal to x digit number in base 2.

So 10^64-1=2^x-1

taking log both side . take log of base 2 so that RHS can become x

64 log10 base2=x

So, x=206.08 

approx 206

0 votes
0 votes
Decimal number 8 and 9 can be represented 4 bits in binary numbers

So, maximum digit to represent a decimal number is 4

To represent 64 bit decimal to binary , number of bits required 64⨉4=256
Answer:

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