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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 14"

(Created page with "==Problem== A rational <math>\frac{1}{k}</math>, where <math>k</math> is a positive integer, is said to be <math>\textit{n-unsound}</math> if its base <math>N</math> represent...")
 
(solution)
 
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==Solution==
 
==Solution==
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{{Solution}}
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Denote the set of distinct primes dividing into <math>k</math> as <math>X</math>. The rational <math>\frac{1}{k}</math> is only terminating in base <math>N</math> if for all members <math>x\in X</math>, we have <math>x|N</math>. As a result, <math>2</math>, <math>4</math>, and <math>8</math> are identical in this respect, for instance, and we can thus reduce the <math>n</math> to:
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<cmath>2,3,5,7,11,13,2\cdot3,2\cdot5,2\cdot7</cmath>
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Clearly for each of the primes <math>p</math> in the list above, the sum of all terminating rationals of such a form is
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<cmath>\sum_{i=0}^\infty\frac{1}{p^i}=\frac{p}{p-1}</cmath>
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Thus the primes <math>2,3,5,7,11,13</math> respectively contribute
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<cmath>2,\frac{3}{2},\frac{5}{4},\frac{7}{6},\frac{11}{10},\frac{13}{12}</cmath>
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to the total sum. However, each of them overcounts the case <math>k=1</math>, so we must subtract <math>5</math> at the end to account for this.
 +
 
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Next, we consider the three <math>n</math> with multiple distinct prime factors. If we ignore all values already counted above, then the sum of each, for <math>p=3,5,7</math>, is,
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<cmath>\sum_{i=1}^\infty\frac{1}{2^i}\cdot\sum_{i=1}^\infty\frac{1}{p^i}=\frac{1}{p-1}</cmath>
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Thus the overall sum is
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<cmath>2+\frac{3}{2}+\frac{5}{4}+\frac{7}{6}+\frac{11}{10}+\frac{13}{12}-5+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{241}{60}</cmath>
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The solution is therefore <math>241+60=\boxed{301}</math>.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]

Latest revision as of 20:33, 29 April 2025

Problem

A rational $\frac{1}{k}$, where $k$ is a positive integer, is said to be $\textit{n-unsound}$ if its base $N$ representation terminates. Let $S_n$ be the set of all $\textit{n-unsound}$ rationals. The sum of all the elements in the union set $S_2\cup S_3\cup\cdots\cup S_{14}$ is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Denote the set of distinct primes dividing into $k$ as $X$. The rational $\frac{1}{k}$ is only terminating in base $N$ if for all members $x\in X$, we have $x|N$. As a result, $2$, $4$, and $8$ are identical in this respect, for instance, and we can thus reduce the $n$ to: \[2,3,5,7,11,13,2\cdot3,2\cdot5,2\cdot7\] Clearly for each of the primes $p$ in the list above, the sum of all terminating rationals of such a form is \[\sum_{i=0}^\infty\frac{1}{p^i}=\frac{p}{p-1}\] Thus the primes $2,3,5,7,11,13$ respectively contribute \[2,\frac{3}{2},\frac{5}{4},\frac{7}{6},\frac{11}{10},\frac{13}{12}\] to the total sum. However, each of them overcounts the case $k=1$, so we must subtract $5$ at the end to account for this.

Next, we consider the three $n$ with multiple distinct prime factors. If we ignore all values already counted above, then the sum of each, for $p=3,5,7$, is, \[\sum_{i=1}^\infty\frac{1}{2^i}\cdot\sum_{i=1}^\infty\frac{1}{p^i}=\frac{1}{p-1}\] Thus the overall sum is \[2+\frac{3}{2}+\frac{5}{4}+\frac{7}{6}+\frac{11}{10}+\frac{13}{12}-5+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{241}{60}\] The solution is therefore $241+60=\boxed{301}$.

~ eevee9406

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