5.2k views

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
| 5.2k views
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I think this question was missing some extra info like log5=xxx and log10 = xxx ? I lost 3 marks and bonus - one negative mark for wrong answer. I took log10 value as 3 and then finally got 192 bits and approximated it as 200 as i was not aware of log10 base 2 value:(I know it would be little more than 3 but not sure how much))

P.S.: My attitude in those days was like why should I memorize things, we are Engineers if needed we should derive it on the fly but I failed in rush of exam,  but if you ask me now I would definitely say memorizing few things will help saving your time in competitive exams like ISRO.  I learned my lesson :) I hope it helps some one :)
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How do we calculate log10base2 in exam without calculator.

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

$10^{64} = 2^x$

$x = 64 . \log_2 10$

$x = 64 \times 3.32 \approx 213.$
by (189 points)
edited by
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i think you are correct but there is a mistake in the last step. The $log_{2}10$ = 3.32 not 3.21 but final answer is correct.
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why u have taken -1 on both the sides in step 1?

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i did not get....
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See now..
+3
maximum value represented by n-digit number is given by ; (Base^n) -1 ...you can check for example 3 bit binary which represent maximum number is 111 whose decimal value equivalent is (2^3)-1=7
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@ponagraj How to solve this type of questions in which log is involved do we need to remember log values.

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

by Boss (38.7k points)
edited by
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how do u create log1o 2?  Go detail

Suppose all are 9

means 64 number of 9 , then how do u represent by 213 bits?

+1

Formula applicable for all decimal number with base 10.

I am editing ans see ?
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@srestha .... decimal is not like hexa-decimal where each digit takes 4 bit ......
even if all 64 digits in decimal are 9 then also its binary representation will have maximum 213 bits.
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yes @Manoj @vijay

I got it

but I think @Manoj u r applying wrong formula

max number of bits in d digit number is floor(log210) +1 =193

I also do this like that

1 bit decimal can be represent in (1 to 4) bit binary

2 bit decimal can be represent in (4 to 7) bit binary

3 bit decimal can be represent in (7 to 10) bit binary

So, n bit decimal can be represent 3⨉n +1 bit binary = 3⨉64+1 =193 bit binary

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

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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

by Boss (13.8k points)
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thankuuuu
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take log value correctly ans=213 bits
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
by Veteran (119k points)
+2
maximum number having n digit with base-x can be expressed as x^n-1

for 64 in base-10 , it will be 10^64-1

let in binary no. of digit is n then number will be 2^n-1

now, 10^64-1<=2^n-1

nlog2 = 64log10

n =64*1/0.3010

hence n=213 so it is (b)