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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, 8 months ago

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2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

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

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

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

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

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

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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

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

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

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

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