Floors

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).

#FloorFunction

Easy Math Editor

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`bold` or `__bold__`	bold
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1. numbered 2. list	numbered list
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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Michael Tang - 7 years ago

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

Cody Johnson - 7 years ago

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

Daniel Liu - 7 years ago

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

Xuming Liang - 7 years ago

@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 - 7 years ago

@Cody Johnson – For negatives, $x\lfloor x\rfloor \geq x^2$ .

Michael Lee - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$