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Difference between revisions of "2023 SSMO Relay Round 5 Problems/Problem 1"

(Created page with "==Problem== Let <math>S_n</math> be the set of all rational numbers of the form <math>0.\overline{a_1a_2a_3\dots a_n},</math> where <math>n</math> is an integer satisfying <ma...")
 
(solution)
 
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==Solution==
 
==Solution==
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To find <math>\sum_{a\in S_n}\frac{a}{10^{n}}</math>, we just find the average of all numbers in <math>S_n</math> and multiply by the size of the set. The average of the nonzero digits is <math>5</math>, so the average of all numbers in <math>S_n</math> is <math>0.\overline{555\dots5}=\frac{5}{9}</math>. There are <math>9</math> choices for each <math>a_i</math>, so there are <math>9^n</math> terms in <math>S_n</math>; thus the sum is <math>\frac{5}{9}\cdot9^n=5\cdot9^{n-1}</math>, so
 +
<cmath>\sum_{a\in S_n}\frac{a}{10^{n}}=\frac{5\cdot9^{n-1}}{10^n}</cmath>
 +
 +
Then the equation becomes
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<cmath> 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</cmath>
 +
By geometric series, <math>\sum_{n=0}^{\infty}\left(\frac{9}{10}\right)^n=\frac{1}{1-\frac{9}{10}}=10</math>, so
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<cmath>n=\frac{5}{2}\cdot 10=\boxed{25}</cmath>
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 +
<math>\textit{Also, a tip for the writer: try to avoid using}</math> <math>n</math> <math>\textit{as both a variable and a dummy variable in the same equation.}</math>
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 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]

Latest revision as of 21:39, 19 March 2025

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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