Question

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

$197$

$198$

$199$

$200$

$201$

Answer: C

Concepts:floor and ceiling functions counting intersections

Difficulty rating: 2500

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 $[k,k+1),$ write $x=k+t$ with $0\le t\lt1.$ The equation becomes $(k+t)^2-10{,}000t=0.$ For $-100\le k\le98,$ the left side is strictly decreasing; at $t=0$ it is $k^2\ge0,$ while as $t$ approaches $1$ it approaches $(k+1)^2-10{,}000\lt0.$ Thus each of these intervals contains exactly one solution. There are $98-(-100)+1=199$ such intervals.

Thus, the correct answer is C.

2018 AMC 12B Problem 24

Answer: C

Difficulty rating: 2500

Solution:

Problem 24 in Other Years