# Why does it take maximum of $n/\log n$ digits to represent the number $2^n - 1$ in base of $n$?

Given the number $n$. Why does it take maximum of $\frac{n}{\log n}$ digits to represent the number $2^n - 1$ in base of $n$?

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Why is this listed as complex analysis? –  Thomas Andrews Dec 13 '12 at 18:51
Thanks Thomas Andrews! –  user1901968 Dec 13 '12 at 18:51
Because it is a part of a complexity algorithm called Radix-Sort –  user1901968 Dec 13 '12 at 18:52
Complexity is not the same as complex analysis, which is analysis on the complex plane. –  Thomas Andrews Dec 13 '12 at 18:53
Changed the tag :) –  user1901968 Dec 13 '12 at 18:54
Hint: to express any number $m$ in base $n$ takes $\lfloor \log_n m \rfloor +1$ digits. To see this, think about base $10$. With one digit you can express any number up to $9$, for which the floor of the log is $0$. With three digits you can express numbers up to $999$, for which the floor is $2$. What base of logs were you planning to use?