2023 SSMO Relay Round 5 Problems/Problem 1

Problem

Let $S_n$ be the set of all rational numbers of the form $0.\overline{a_1a_2a_3\dots a_n},$ where $n$ is an integer satisfying $n\geq 1$ and $a_1,a_2,\dots,a_n$ are nonzero integers. If $n = 5\left(\sum_{n=1}^{\infty}\left(\sum_{a\in S_n}\frac{a}{10^{n}}\right)\right),$ find $n.$

Solution

To find $\sum_{a\in S_n}\frac{a}{10^{n}}$ , we just find the average of all numbers in $S_n$ and multiply by the size of the set. The average of the nonzero digits is $5$ , so the average of all numbers in $S_n$ is $0.\overline{555\dots5}=\frac{5}{9}$ . There are $9$ choices for each $a_i$ , so there are $9^n$ terms in $S_n$ ; thus the sum is $\frac{5}{9}\cdot9^n=5\cdot9^{n-1}$ , so $\sum_{a\in S_n}\frac{a}{10^{n}}=\frac{5\cdot9^{n-1}}{10^n}$

Then the equation becomes $n = 5\left(\sum_{n=1}^{\infty}\left(\sum_{a\in S_n}\frac{a}{10^{n}}\right)\right)=5\left(\sum_{n=1}^{\infty}\frac{5\cdot9^{n-1}}{10^n}\right)=\frac{25}{10}\left(\sum_{n=1}^{\infty}\frac{9^{n-1}}{10^{n-1}}\right)=\frac{5}{2}\sum_{n=0}^{\infty}\left(\frac{9}{10}\right)^n$ By geometric series, $\sum_{n=0}^{\infty}\left(\frac{9}{10}\right)^n=\frac{1}{1-\frac{9}{10}}=10$ , so $n=\frac{5}{2}\cdot 10=\boxed{25}$

$\textit{Also, a tip for the writer: try to avoid using}$ $n$ $\textit{as both a variable and a dummy variable in the same equation.}$

~ eevee9406

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2023 SSMO Relay Round 5 Problems/Problem 1

Problem

Solution