AoPS Wiki talk:Problem of the Day/June 10, 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 $\boxed{3}$ .

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AoPS Wiki talk:Problem of the Day/June 10, 2011

Problem

Solution