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

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<math>\linebreak</math>
 
<math>\linebreak</math>
 
~Apple321
 
~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]]
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{{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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