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# 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).

Note by Cody Johnson
2 years, 12 months ago

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Is G guts? My first reaction was "why is this a geometry problem!?" :P · 2 years, 12 months ago

Step One: Let $$x=q+r$$ where $$q\in\mathbb{Z}$$ and $$0\le r<1$$. · 2 years, 12 months ago

It is true that either $$q=1$$ or $$q=2$$. We can quickly find the answer afterwards. · 2 years, 12 months ago

Yeah we can also bound it like this: $$2x^2-5=x\lfloor x\rfloor\le x^2$$ · 2 years, 12 months ago

Can this be applied to negative numbers too? Why is $$-\frac52$$ a solution, yet $$\left(-\frac52\right)^2=6.25>5$$? · 2 years, 11 months ago

For negatives, $$x\lfloor x\rfloor \geq x^2$$. · 2 years, 11 months ago