The Shorter The Better

Computer Science Level 3

Alice and Bob wants to send a binary number (base 2) with 2400 digits to Chris. However, the internet speed is ridiculously slow and transmit such a long number can take forever. In order to shorten the length of the number, each of them propose a plan :

Alice: we should convert the number to base 64.
Bob: we should convert the number to base 16.

Let the number generated by Alice and Bob be $A$ and $B$ respectively, and $d(x)$ denote the number of digits in $x$ .

Which option best describe the value of $\dfrac{d(A)}{ d(B)}$ ?

\frac14

\frac31

\frac32

\frac13

\frac43

\frac23

\frac34

\frac41

1 solution

Agnishom Chattopadhyay
Sep 2, 2016

A way to think about this is to notice that to translate a number from binary to hexadecimal, 4 bits collapse into a single hexadecimal digit. But, to translate a number from binary to base-64, 6 bits collapse into a single digit.

To see why, notice that $4$ bits describe $16$ distinct numbers, which can be done with just one hexadecimal digit. And similarly for base-64 numerals

This is why $\frac{4}{6}$ times the space is needed.

The number of different things (read numbers ) that one can describe with 2400 binary digits is $2^{2400}$ .

To describe the same number of things using hexadecimal digits, we just need $log_{16}2^{2400}$ . And similarly, with base 64, we need $log_{64}2^{2400}$

So, the answer is $\frac{\log_{64}2^{2400}}{\log_{16}2^{2400}}= \frac{\log_{64} 2}{\log_{16} 2} = \frac{\frac{1}{6} \log 2}{\frac{1}{4} \log 2} = \frac{2}{3}$

Agnishom Chattopadhyay - 4 years, 9 months ago

The Shorter The Better

1 solution

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