# How many bits needed to store a number

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

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

-
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 –  Ask Jun 19 '12 at 11:58
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.