2021 JMPSC Accuracy Problems/Problem 9


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 $x_1+x_2+\cdots+x_{10}$.


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 $\frac{12(13)}{2}-3=\boxed{75}$


See also

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

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