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

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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)
 
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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== Alternative, Bashy Solution (Please excuse me for bad latex) ==
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== Alternative, Bashy Solution ==
  
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Same as the first solution, x^2=10,000{x}.
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Same as the first solution, <math>x^2=10,000\{x\} </math>.
  
  
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We can write x as floor(x)+{x}. Expanding everything, we get a quadratic in {x} in terms of floor(x):
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We can write <math>x</math> as <math>\lfloor x \rfloor+\{x\}</math>. Expanding everything, we get a quadratic in <math>{x}</math> in terms of <math>\lfloor x \rfloor</math>:
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{x}^2+ (2*floor(x)-10,000){x} + [floor(x)]^2 = 0
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<math>\{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0</math>
  
  
 
We use the quadratic formula to solve for {x}:
 
We use the quadratic formula to solve for {x}:
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{x} = [ -2*floor(x)+10,000 +/- sqrt( [-2*floor(x)+10,000]^2- 4*[floor(x)]^2 ) ]/2
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<math>\{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{( -2 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 ) }}{2} </math>
  
  
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Since 0=<{x}<1, we get an inequality which we can then solve. After simplifying a lot, we get that [floor(x)]^2 + 2*floor(x) - 9999 < 0
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Since <math> 0 \leq \{x\} < 1 </math>, we get an inequality which we can then solve. After simplifying a lot, we get that <math>\lfloor x \rfloor^2 + 2\lfloor x \rfloor - 9999 < 0</math>.
  
  
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Solving over the integers, -101< floor(x) < 99, and since floor(x) is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of floor(x) should correspond to one value of x, so we are done.
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Solving over the integers, <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of<math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done.
  
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~Alex_z_Awesome
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~ Alex_z_Awesome
  
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*This solution needs some editing. Please edit if you have time.
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edited by <math> \tiny{math4fun2} pt\par</math>
  
 
==See Also==
 
==See Also==

Revision as of 19:02, 17 February 2018

Problem

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

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

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 hole 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)

Alternative, Bashy Solution

Same as the first solution, $x^2=10,000\{x\}$.


We can write $x$ as $\lfloor x \rfloor+\{x\}$. Expanding everything, we get a quadratic in ${x}$ in terms of $\lfloor x \rfloor$: $\{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0$


We use the quadratic formula to solve for {x}: $\{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{( -2 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 ) }}{2}$


Since $0 \leq \{x\} < 1$, we get an inequality which we can then solve. After simplifying a lot, we get that $\lfloor x \rfloor^2 + 2\lfloor x \rfloor - 9999 < 0$.


Solving over the integers, $-101 < \lfloor x \rfloor < 99$, and since $\lfloor x \rfloor$ is an integer, there are $\boxed{\text{(C)}~199}$ solutions. Each value of$\lfloor x \rfloor$ should correspond to one value of $x$, so we are done.

~ Alex_z_Awesome

edited by $\tiny{math4fun2} pt\par$ (Error compiling LaTeX. Unknown error_msg)

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

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