2021 JMPSC Accuracy Problems/Problem 9

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

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png