# 2021 April MIMC 10 Problems/Problem 19

$(0.51515151...)_n$ can be expressed as $(\frac{6}{n})$ in base $10$ which $n$ is a positive integer. Find the sum of the digits of $n^{3}$.

$\textbf{(A)} ~6 \qquad\textbf{(B)} ~7 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~9 \qquad \textbf{(E)} ~\textrm{Does Not Exist}$

## Solution

We realize that when a decimal $0.abcdef$ is expressed in base $n$, the decimal would equal to the expression $\frac{a}{n}+\frac{b}{n^2}+\frac{c}{n^3}+\frac{d}{n^4}+\frac{e}{n^5}+\frac{f}{n^6}+......$. Use this idea, $(0.515151515151.....)_n=\frac{5}{n}+\frac{1}{n^2}+\frac{5}{n^3}+\frac{1}{n^4}+\frac{5}{n^5}+\frac{1}{n^6}+....$.

This sum is basically the sum of two infinite geometric series. The first one has first term of $\frac{5}{n}$ and a common ratio of $\frac{1}{n^2}$. The second one has first term $\frac{1}{n^2}$ and a common ratio of $\frac{1}{n^2}$. The total sum is $\frac{5n+1}{n^2-1}$. This would result in $\frac{5n+1}{n^2-1}=\frac{6}{n}$. Turn this into a quadratic by cross-multiplication, we would get $n=3$. HOWEVER, all numbers in base $3$ can only have $0,1,2$ as its digits. Therefore, the answer will be $\fbox{\textbf{(E)} Does Not Exist}$.