2021 JMPSC Accuracy Problems/Problem 9: Difference between revisions

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

@@ Line 20: / Line 20: @@
 ==See also==
-#[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]
+#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
-#[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]
+#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
 #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 {{JMPSC Notice}}

Page

Toolbox

Search

2021 JMPSC Accuracy Problems/Problem 9: Difference between revisions

Latest revision as of 16:24, 11 July 2021

Contents

Problem

Solution

Solution 2

See also