2
$\begingroup$

Given some number $N$ of $d$ digits in base and $b > 1$ such that - $b^{d-1} \le N < b^d$.

Prove that any number of $d$ digits would be of $\lceil dlog_cb\rceil$ digits at most at some base $c$.

My Attempt -

$b^{d-1} \le N <b^d \iff log_cb^{d-1} \le log_cN < log_cb^d \iff (d-1)log_cb \le log_cN < dlog_cb$

It is also known that the number of digits for $N$ in base $c$ is - $\lfloor log_cN\rfloor +1.$ I think i might need to use it but am not sure how to proceed that way.

Thank you !

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.