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Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 9"

(Created page with "==Problem== If <math>x_1,x_2,\ldots,x_{10}</math> is a strictly increasing sequence of positive integers that satisfies <cmath>\frac{1}{2}<\frac{2}{x_1}<\frac{3}{x_2}< \cdots...")
 
 
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==Solution==
 
==Solution==
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asdf
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Say we take <math>x_1,x_1,x_3,...,x_{10}</math> as <math>4,5,6,...,13</math> as an example. The first few terms of the inequality would then be:
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<cmath>\frac{1}{2}<\frac{2}{4}<\frac{3}{5}<\frac{4}{6}</cmath>
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But <math>\frac{3}{5}<\frac{4}{6}</math>, reaching a contradiction.
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A contradiction will also be reached at some point when <math>x_1\geq 4</math> or when <math>x_1\leq 2</math>, so that must mean <math>x_1=3</math>.
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<math>\implies 3+4+5+...+12=\frac{10\cdot 15}{2}=\boxed{75}</math>
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<math>\linebreak</math>
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~Apple321
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==Solution 2==
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We recall the identity that <math>\frac{x}{x+1}</math> is monotically increasing. Here, we have the same case, <math>x_1=3</math>, <math>x_2=4</math>, and so on. The answer is <math>\frac{12(13)}{2}-3=\boxed{75}</math>
 +
 
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~Geometry285
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==See also==
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#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
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#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
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#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 +
{{JMPSC Notice}}

Latest revision as of 16:24, 11 July 2021

Problem

If $x_1,x_2,\ldots,x_{10}$ is a strictly increasing sequence of positive integers that satisfies \[\frac{1}{2}<\frac{2}{x_1}<\frac{3}{x_2}< \cdots < \frac{11}{x_{10}},\] find $x_1+x_2+\cdots+x_{10}$.

Solution

Say we take $x_1,x_1,x_3,...,x_{10}$ as $4,5,6,...,13$ as an example. The first few terms of the inequality would then be: \[\frac{1}{2}<\frac{2}{4}<\frac{3}{5}<\frac{4}{6}\] But $\frac{3}{5}<\frac{4}{6}$, reaching a contradiction.

A contradiction will also be reached at some point when $x_1\geq 4$ or when $x_1\leq 2$, so that must mean $x_1=3$.

$\implies 3+4+5+...+12=\frac{10\cdot 15}{2}=\boxed{75}$ $\linebreak$ ~Apple321

Solution 2

We recall the identity that $\frac{x}{x+1}$ is monotically increasing. Here, we have the same case, $x_1=3$, $x_2=4$, and so on. The answer is $\frac{12(13)}{2}-3=\boxed{75}$

~Geometry285

See also

  1. Other 2021 JMPSC Accuracy Problems
  2. 2021 JMPSC Accuracy Answer Key
  3. All JMPSC Problems and Solutions

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