A Beautiful Problem on Base

Number Theory Level 2

I have a problem which I would like to share and also I want to be satisfied with the answer I have found.The problem is-

Let $p(n)$ be the number of digits when $8^n$ is written in base $6$ and let $q(n)$ is the number of digits when $6^n$ is written in base $4$ .For example $8^2$ in base $6$ is $144$ .hence $p(2)=3$ . Then find the value of $\lim_{n \to\ \infty} \frac{{p(n)}{q(n)}}{n^2}$

3/2 1 4/3 2

1 solution

Alapan Das
Dec 14, 2018

8^n=(6+2)^n=6^n(4/3)^n and for 6^n in base 6 there are n+1 digits. Now if 6^(r+1)> (4/3)^n>6^(r) then there are n+r+1 digits in base 6.So,r<(n (ln4/3)/(ln6)).Hence p(n)=n+1+(n (ln4/3)/(ln6)).Same as this q(n)=n+1+(n*(ln3/2)/(ln4)).So, lim┬(n→∞)⁡〖(p(n)q(n))/n^2 〗=lim┬(n→∞)⁡(1+ln(4/3)/ln6)(1+ln(3/2)/ln4)=3/2

Nice, but you need to upgrade to windows. Das is not used anymore.

A Former Brilliant Member - 2 years, 5 months ago

A Beautiful Problem on Base

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