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How many bits needed to store a number $55^{2002}$ ?

My answer is $2002\;\log_2(55)$, is it correct?

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That would be bits. Also you need a ceiling function in the end. – Jyrki Lahtonen Jun 19 '12 at 11:57
yeah, I meant bits, sorry – user1192466 Jun 19 '12 at 11:58
00110101 00110101 01011110 00110010 00110000 00110000 00110010 are only 56 bits :) – Hagen von Eitzen Jan 27 '14 at 16:48
@HagenvonEitzen: It says number, not string. And fact you can store ASCII at 7 bits per character.;-) – Marc van Leeuwen Jan 27 '14 at 17:52
@MarcvanLeeuwen In order to be able to express more, I used UTF-8 ;) - But the answer to the original question should nevertheless depend on the encoding to be precise. For example IEEE floats are very well suited to store numbers that are powers of two, or (relatively) small numbers times a power of two. So who stops us from introducing an encoding that is good (and exact) for powers of $55$? – Hagen von Eitzen Jan 28 '14 at 19:31
up vote 5 down vote accepted

The number of bits required to represent an integer $n$ is $\lfloor\log_2 n\rfloor+1$, so $55^{2002}$ will require $\lfloor 2002\; \log_2 55\rfloor+1$ bits, which is $11,575$ bits.

Added: For example, the $4$-bit integers are $8$ through $15$, whose logs base $2$ are all in the interval $[3,4)$. We have $\lfloor\log_2 n\rfloor=k$ if and only if $k\le\log_2 n<k+1$ if and only if $2^k\le n<2^{k+1}$, and that’s exactly the range of integers requiring $k+1$ bits.

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