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Difference between revisions of "2018 AMC 10B Problems/Problem 25"

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== Problem ==
 
How many <math>x</math> satisfy the equation <math>x^2 + 10,000\lfloor x \rfloor = 10,000x</math>?
 
How many <math>x</math> satisfy the equation <math>x^2 + 10,000\lfloor x \rfloor = 10,000x</math>?
  
 
<math>\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201</math>
 
<math>\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201</math>
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== Solution ==
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This rewrites itself to <math>x^2=10,000\{x\}</math>.
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Graphing <math>y=10,000\{x\}</math> and <math>y=x^2</math> we see that the former is a set of line segments with slope <math>10,000</math> from <math>0</math> to <math>1</math> with a hole at <math>x=1</math>, then <math>1</math> to <math>2</math> with a whole at <math>x=2</math> etc.
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Here is a graph of <math>y=x^2</math> and <math>y=16\{x\}</math> for visualization.
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<asy>
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import graph;
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size(400);
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xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5}));
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yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}));
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real y(real x) {return x^2;}
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draw(circle((-4,16), 0.1));
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draw(circle((-3,16), 0.1));
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draw(circle((-2,16), 0.1));
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draw(circle((-1,16), 0.1));
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draw(circle((0,16), 0.1));
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draw(circle((1,16), 0.1));
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draw(circle((2,16), 0.1));
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draw(circle((3,16), 0.1));
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draw(circle((4,16), 0.1));
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draw((-5,0)--(-4,16), black);
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draw((-4,0)--(-3,16), black);
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draw((-3,0)--(-2,16), black);
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draw((-2,0)--(-1,16), black);
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draw((-1,0)--(-0,16), black);
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draw((0,0)--(1,16), black);
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draw((1,0)--(2,16), black);
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draw((2,0)--(3,16), black);
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draw((3,0)--(4,16), black);
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draw(graph(y,-4.2,4.2),green);
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</asy>
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Now notice that when <math>x=\pm 100</math> then graph has a hole at <math>(\pm 100,10,000)</math> which the equation <math>y=x^2</math> passes through and then continues upwards. Thus our set of possible solutions is bounded by <math>(-100,100)</math>. We can see that <math>y=x^2</math> intersects each of the lines once and there are <math>99-(-99)+1=199</math> lines for an answer of <math>\boxed{\text{(C)}~199}</math>. (Mudkipswims42)
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==See Also==
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{{AMC10 box|year=2018|ab=B|num-b=24|after=Last Problem}}
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{{AMC12 box|year=2018|ab=B|num-b=23|num-a=25}}
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{{MAA Notice}}

Revision as of 13:41, 16 February 2018

Problem

How many $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$?

$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$

Solution

This rewrites itself to $x^2=10,000\{x\}$.

Graphing $y=10,000\{x\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with a hole at $x=1$, then $1$ to $2$ with a whole at $x=2$ etc.

Here is a graph of $y=x^2$ and $y=16\{x\}$ for visualization.

[asy] import graph; size(400); xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5})); yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18})); real y(real x) {return x^2;} draw(circle((-4,16), 0.1)); draw(circle((-3,16), 0.1)); draw(circle((-2,16), 0.1)); draw(circle((-1,16), 0.1)); draw(circle((0,16), 0.1)); draw(circle((1,16), 0.1)); draw(circle((2,16), 0.1)); draw(circle((3,16), 0.1)); draw(circle((4,16), 0.1)); draw((-5,0)--(-4,16), black); draw((-4,0)--(-3,16), black); draw((-3,0)--(-2,16), black); draw((-2,0)--(-1,16), black); draw((-1,0)--(-0,16), black); draw((0,0)--(1,16), black); draw((1,0)--(2,16), black); draw((2,0)--(3,16), black); draw((3,0)--(4,16), black); draw(graph(y,-4.2,4.2),green); [/asy]

Now notice that when $x=\pm 100$ then graph has a hole at $(\pm 100,10,000)$ which the equation $y=x^2$ passes through and then continues upwards. Thus our set of possible solutions is bounded by $(-100,100)$. We can see that $y=x^2$ intersects each of the lines once and there are $99-(-99)+1=199$ lines for an answer of $\boxed{\text{(C)}~199}$. (Mudkipswims42)

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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