2018 AMC 12B Problem 24

Below is the professionally curated solution for Problem 24 of the 2018 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12B solutions, or check the answer key.

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Concepts:floor and ceiling functions counting intersections

Difficulty rating: 2500

24.

Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x.$ How many real numbers $x$ satisfy the equation $x^2+10{,}000\lfloor x\rfloor=10{,}000x?$

$197$

$198$

$199$

$200$

$201$

Solution:

Let $\{x\}=x-\lfloor x\rfloor.$ The equation becomes $x^2=10{,}000\{x\},$ so $\tfrac{x^2}{10{,}000}=\{x\}.$ Since $0\le\{x\}\lt1,$ we need $0\le x^2\lt10{,}000,$ i.e. $-100\lt x\lt100.$

On each interval $[k,k+1)$ the increasing parabola $\tfrac{x^2}{10{,}000}$ meets the segment $\{x\}$ exactly once. These intervals run for $k=-100,-99,\ldots,98,$ giving $199$ solutions.

Thus, the correct answer is C.

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