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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>.
 
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>.
  
−
Note: from the graph we can clearly see there are 4 solution on the negative side and only 2 on the positive side. So the solution really should be from -100 to 98, which still counts to 199. A couple of the alternative solutions also seem to have the same flaw.
+
Note: From the graph, we can clearly see there are <math>4</math> solutions on the negative side of the <math>x</math>-axis and only <math>2</math> on the positive side of the <math>x</math>-axis. So the solution really should be from <math>-100</math> to <math>98</math>, which still counts to <math>199</math>. A couple of the alternative solutions also seem to have the same flaw.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 21:00, 9 November 2019

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 1

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 $\boxed{\text{(C)}~199}$.

Note: From the graph, we can clearly see there are $4$ solutions on the negative side of the $x$-axis and only $2$ on the positive side of the $x$-axis. So the solution really should be from $-100$ to $98$, which still counts to $199$. A couple of the alternative solutions also seem to have the same flaw.

Solution 2

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 $\{x\}$ : \[\{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ ( 2\lfloor x \rfloor - 10,000 )^2 - 4\lfloor x \rfloor^2 }}{2} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ 4\lfloor x \rfloor^2 -20,000 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 }}{2}\]


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.

Solution 3

Let $x = a+k$ where $a$ is the integer part of $x$ and $k$ is the fractional part of $x$. We can then rewrite the problem below:

$(a+k)^2 + 10000a = 10000(a+k)$

From here, we get

$(a+k)^2 + 10000a = 10000a + 10000k$

Solving for $a+k = x$

$(a+k)^2 = 10000k$

$x = a+k = \pm100\sqrt{k}$

Because $0 \leq k < 1$, we know that $a+k$ cannot be less than or equal to $-100$ nor greater than or equal to $100$. Therefore:

$-99 \leq x \leq 99$

There are $199$ elements in this range, so the answer is $\boxed{\textbf{(C)} \text{ 199}}$.

Solution 4

Notice the given equation is equivilent to $(\lfloor x \rfloor+\{x\})^2=10,000\{x\}$

Now we now that $\{x\} < 1$ so plugging in $1$ for $\{x\}$ we can find the upper and lower bounds for the values.

$(\lfloor x \rfloor +1)^2 = 10,000(1)$

$(\lfloor x \rfloor +1) = \pm 100$

$\lfloor x \rfloor = 99, -101$

And just like Solution 2, we see that $-101 < \lfloor x \rfloor < 99$, 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.

Solution 5

First, we can let $\{x\} = b, \lfloor x \rfloor = a$. We know that $a + b = x$ by definition. We can rearrange the equation to obtain

$x^2 = 10^4(x - a)$.

By taking square root on both sides, we obtain $x = \pm 100 \sqrt{b}$ (because $x - a = b$). We know since $b$ is the fractional part of $x$, it must be that $0 \leq b < 1$. Thus, $x$ may take any value in the interval $-100 < x < 100$. Hence, we know that there are $\boxed{\text{(C)}~199}$ potential values for $\lfloor x \rfloor$ in that range and we are done.

~awesome1st

Video Solution

https://www.youtube.com/watch?v=vHKPbaXwJUE

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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