NIELIT 2016 MAR Scientist B - Section C: 5

Question

A decimal number has $30$ digits. Approximately, how many digits would the binary representation have?

• A. $30$
• B. $60$
• C. $90$
• D. $120$

Comments

Actually $30 \times log_210$ is $99.66$ approximately, for which ceiling is $100$. $90$ cannot be the answer. The only number greater than $100$ is $120$.

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Maximum 30 digit decimal number = 10^30 -1

Maximum n bit binary number = 2^n - 1

10^30 - 1 = 2^n - 1

n = 30log 10

n = 30*(3.219280948873626)

= 99.657 (approx)

None of the option is matching.

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@vg653 Binary representation have minimum 100 digits(bits).

can we choose option D? @Lakshman Patel RJIT

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Exact answer is 30 *( log 10 base 2)

But in question say that give approximately answers:

It's greater than 90

So I think we choose option D is correct.

But I am again say it's not exact answer.

Answer 1

actually 10³⁰-1 but its close to 10³⁰ so yes you are right , its greater than 90 

so ans will be D)120

Comments

.y yes thats a mistake it should be 10^30-1 .actually i didnt find 96 in the option and there is a large gap in b/w 96 and 120 (specially when calculating the no of digit as each binart digit has a capability to double the no ) thats why i thoought may be i m making a mistake

.......thank u 4 ur help

Answer 2

Maximum value of a $30$ digit number $ = \underbrace{9999\ldots9}_{\text{30 9's}}\approx 10^{30}$

Now,  $10^{30} \leq 2^n$

Taking $\log$ on both sides, we get $\displaystyle 30  \leq n \log 2 \implies n \geq \lceil \dfrac{30}{\log 2}\rceil = 100.$

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Minimum value of a $30$ digit number $ = \underbrace{100\ldots0}_{\text{29 0's}}= 10^{29}$

Now,  $10^{29} \leq 2^n$

Taking $\log$ on both sides, we get $29  \leq n \log 2 \implies n \geq \lceil \dfrac{29}{\log 2}\rceil = 97.$

So, depending on the $30$ digit number, the binary representation will need minimum $97$ digits and maximum $100$ digits.

Best option, C.