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

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 $\lfloor x \rfloor = x-x^2/10,000$

Rearranging, we get $x^2/10,000 < 1$

Noticing that $10,000 = 100^2$, we know that x can only be within the boundaries of $-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, then we know the following:

$0

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

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

Therefore, $0, 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, then we know the following:

$1

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

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

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 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, but not for $100. Similarly, there is a solution for $-100, but not for $-101. This means that there are $99$ positive solutions, $99$ negative solutions, and $0$ for a total of $\boxed{\text{(C)}~199}$ solutions.

~Owen1204