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

Revision as of 18:14, 4 May 2020

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

Note (not by author): this solution seems to be invalid at first, because one can not determine whether $x$ is an integer or not. However, it actually works because although $x$ itself might not be an integer, it is very close to one, so there are 199 potential $x$ .

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 $\textbf{Solution 2}$ , 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

Solution 6

Firstly, we can rearrange to get $\lfloor x \rfloor = x-\frac{x^2}{10,000}$

Rearranging, we get $\frac{x^2}{10,000} < 1$

Noticing that $10,000 = 100^2$ , we know that x can only be within the boundaries of $-100<x<100$ and hence, we know that there are $\boxed{\text{(C)}~199}$ potential values.

Solution 7

Firstly, if $x$ is an integer, then $10,000\lfloor x \rfloor=10,000x$ , so $x$ must be $0$ .

If $0<x<1$ , then we know the following:

$0<x^2<1$

$10,000\lfloor x \rfloor =0$

$0<10,000x<10,000$

Therefore, $0<x^2+10,000\lfloor x \rfloor <1$ , which overlaps with $0<10,000x<10,000$ . This means that there is at least one real solution between $0$ and $1$ . Since $x^2+10,000\lfloor x \rfloor$ increases exponentially and $10,000x$ increases linearly, there is only one solution for this case.

Similarly, if $1<x<2$ , then we know the following:

$1<x^2<4$

$10,000\lfloor x \rfloor =1$

$<10,000<10,000x<20,000$

By following similar logic, we can find that there is one solution between $1$ ad $2$ .

We can also follow the same process to find that there are negative solutions for $x$ as well.

There are not an infinite amount of solutions, so at one point there will be no solutions when $n<x<n+1$ for some positive integer $n$ . Looking at the answer solutions, they are near $200$ , so it seems logical that the upper and lower limits are around $100$ and $-100$ . Testing this out, we can see that there is a solution for $99<x<100$ , but not for $100<x<101$ . Similarly, there is a solution for $-100<x<-99$ , but not for $-101<x<-100$ . This means that there are $99$ positive solutions, $99$ negative solutions, and $0$ for a total of $\boxed{\text{(C)}~199}$ solutions.

~Owen1204

Video Solution

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

@@ Line 110: / Line 110: @@
 ==Solution 6==
-Firstly, we can rearrange to get <math>\lfloor x \rfloor = x-x^2/10,000</math>
+Firstly, we can rearrange to get <math>\lfloor x \rfloor = x-\frac{x^2}{10,000}</math>
-Rearranging, we get <math>x^2/10,000 < 1</math>
+Rearranging, we get <math>\frac{x^2}{10,000} < 1</math>
 Noticing that <math>10,000 = 100^2</math>, we know that x can only be within the boundaries of <math>-100<x<100</math> and hence, we know that there are <math>\boxed{\text{(C)}~199}</math> potential values.

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

Page

Toolbox

Search