Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 10, 2011"

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{{:AoPSWiki:Problem of the Day/June 10, 2011}}
 
==Solution==
 
==Solution==
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<math>5x = \lfloor x + 1 \rfloor + \lceil x + 2 \rceil</math>, <math>0 < x < 2</math>.
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If <math>x</math> is an integer, <math>5x = x + 1 + x + 2 \implies x = 1</math>.
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If <math>x</math> isn't an integer, <math>5x = \lfloor x + 1 \rfloor + \lceil x + 2 \rceil = \lfloor x \rfloor + 1 + \lfloor x \rfloor + 2 + 1 \implies 5x = 2 \lfloor x \rfloor + 4</math>.
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Since <math>\lfloor x \rfloor = 0</math> or <math>1</math>, <math>5x = 4 \implies x = \dfrac{4}{5}</math> or <math>5x = 6 \implies x = \dfrac{6}{5}</math>. Checking all our solutions, they all work. Thus the solutions are <math>x = \dfrac{4}{5}</math>, <math>1</math>, or <math>\dfrac{6}{5}</math>, and the sum of them all is <math>\boxed{3}</math>.

Latest revision as of 22:25, 9 June 2011

Problem

AoPSWiki:Problem of the Day/June 10, 2011

Solution

$5x = \lfloor x + 1 \rfloor + \lceil x + 2 \rceil$, $0 < x < 2$. If $x$ is an integer, $5x = x + 1 + x + 2 \implies x = 1$. If $x$ isn't an integer, $5x = \lfloor x + 1 \rfloor + \lceil x + 2 \rceil = \lfloor x \rfloor + 1 + \lfloor x \rfloor + 2 + 1 \implies 5x = 2 \lfloor x \rfloor + 4$. Since $\lfloor x \rfloor = 0$ or $1$, $5x = 4 \implies x = \dfrac{4}{5}$ or $5x = 6 \implies x = \dfrac{6}{5}$. Checking all our solutions, they all work. Thus the solutions are $x = \dfrac{4}{5}$, $1$, or $\dfrac{6}{5}$, and the sum of them all is $\boxed{3}$.