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This is a seemingly simple problem that I found very interesting, with a surprising answer. Find the sum of all positive solutions to \(2x^2-x\lfloor x\rfloor=5\) (HMNT 2011 G5).

Note by Cody Johnson
3 years, 5 months ago

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Is G guts? My first reaction was "why is this a geometry problem!?" :P

Michael Tang - 3 years, 5 months ago

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Step One: Let \(x=q+r\) where \(q\in\mathbb{Z}\) and \(0\le r<1\).

Cody Johnson - 3 years, 5 months ago

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It is true that either \(q=1\) or \(q=2\). We can quickly find the answer afterwards.

Daniel Liu - 3 years, 5 months ago

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Yeah we can also bound it like this: \(2x^2-5=x\lfloor x\rfloor\le x^2\)

Xuming Liang - 3 years, 5 months ago

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@Xuming Liang Can this be applied to negative numbers too? Why is \(-\frac52\) a solution, yet \(\left(-\frac52\right)^2=6.25>5\)?

Cody Johnson - 3 years, 5 months ago

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@Cody Johnson For negatives, \(x\lfloor x\rfloor \geq x^2\).

Michael Lee - 3 years, 5 months ago

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