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

Latest revision as of 11:51, 29 January 2024

The following problem is from both the 2018 AMC 10B #25 and 2018 AMC 12B #24, so both problems redirect to this page.

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 $x^2=10,000\{x\}$ where $\lfloor x \rfloor + \{x\} = x$ .

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 $x=\pm 100$ the 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 $\boxed{\text{(C)}~199}$ .

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 $\{x\}$ : $\{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ ( 2\lfloor x \rfloor - 10,000 )^2 - 4\lfloor x \rfloor^2 }}{2} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ 4\lfloor x \rfloor^2 -40,000 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 }}{2}$

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 $x$ .

Another Note (not by author of previous note): we can actually determine that $x$ =0 is the only possible integer value of $x$ is we set $x$ = $\lfloor x \rfloor$ we end up with $x$ =0 ~YJC64002776

Solution 4

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

Now we know 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

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 quadratically and $10,000x$ increases linearly, there is only one solution for this case.

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

$1<x^2<4$

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

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

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 integer $n$ . For there to be no solutions in a given range means that the range of $10,000\lfloor x \rfloor + x^2$ does not intersect the range of $10,000x$ . $x^2$ will always be positive, and $10,000\lfloor x \rfloor$ is less than $10,000$ less than $10,000x$ , so when $x^2 >= 10,000$ , the equation will have no solutions. This means that there are $99$ positive solutions, $99$ negative solutions, and $0$ for a total of $\boxed{\text{(C)}~199}$ solutions.

~Owen1204

Solution 6 (General Equation)

General solution to this type of equation $f(x, \lfloor x \rfloor) = 0$ :

1. solve  for  to get 
2. apply , solve  to get the domain of 
3. get  from the domain of  because  is integer, then get  from  by 
Note: function  maps  to its floor. By solving , we get function , mapping 's floor to

$x^2 - 10000x + 10000 \lfloor x \rfloor =0$

$x=5000 \pm 100 \sqrt{2500- \lfloor x \rfloor}$ , $\lfloor x \rfloor \le 2500$

$\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$

If $x= 5000 + 100 \sqrt{2500 - \lfloor x \rfloor}$ , $x \ge 5000$ , it contradicts $x < \lfloor x \rfloor + 1 \le 2501$

So $x= 5000 - 100 \sqrt{2500 - \lfloor x \rfloor}$

Let $k = \lfloor x \rfloor$ , $x= 5000 - 100 \sqrt{2500 - k}$

$k \le 5000 - 100 \sqrt{2500 - k} < k + 1$

$0 \le 5000 - k - 100 \sqrt{2500 - k} < 1$

$0 \le 2500 - k - 100 \sqrt{2500 - k} + 2500 < 1$

$0 \le (\sqrt{2500 - k} - 50)^2 < 1$

$-1 < \sqrt{2500 - k} - 50 < 1$

$49 < \sqrt{2500 - k} < 51$

$-101 < k < 99$

So the number of $k$ 's values is $99-(-101)-1=199$ . Because $x=5000-100\sqrt{2500-k}$ , for each value of $k$ , there is a value for $x$ . The answer is $\boxed{\textbf{(C)} 199}$

~isabelchen

Solution 7

Subtracting $10000\lfloor x\rfloor$ from both sides gives $x^2=10000(x-\lfloor x\rfloor)=10000\{x\}$ . Dividing both sides by $10000$ gives $\left(\frac{x}{100}\right)^2=\{x\}<1$ . $\left(\frac{x}{100}\right)^2<1$ when $-100<x<100$ so the answer is $\boxed{199}$ .

~randomdude10807

Video Solution

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

@@ Line 1: / Line 1: @@
-{{duplicate|[[2018 AMC 10B Problems|2018 AMC 10B #25]] and [[2018 AMC 12B Problems|2018 AMC 12B #24]]}}
+{{duplicate|[[2018 AMC 10B Problems#Problem 25 | 2018 AMC 10B #25]] and [[2018 AMC 12B Problems#Problem 24|2018 AMC 12B #24]]}}
 == Problem ==
@@ Line 7: / Line 7: @@
 ==Solution 1==
-This rewrites itself to <math>x^2=10,000\{x\}</math>.
+This rewrites itself to <math>x^2=10,000\{x\}</math> where <math>\lfloor x \rfloor + \{x\} = x</math>.
 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 hole at <math>x=2</math> etc.
@@ Line 39: / Line 39: @@
 </asy>
-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> the 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 <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.
-Other Note: The function can also be modeled as <math>\left(\dfrac{x}{100}\right)^2 = \{x\}</math>.
 ==Solution 2==
@@ Line 87: / Line 83: @@
 Note (not by author): this solution seems to be invalid at first, because one can not determine whether <math>x</math> is an integer or not. However, it actually works because although <math>x</math> itself might not be an integer, it is very close to one, so there are 199 potential <math>x</math>.
+Another Note (not by author of previous note): we can actually determine that <math>x</math>=0 is the only possible integer value of <math>x</math> is we set <math>x</math>=<math>\lfloor x \rfloor</math> we end up with <math>x</math>=0   ~YJC64002776
 ==Solution 4==
-Notice the given equation is equivilent to <math>(\lfloor x \rfloor+\{x\})^2=10,000\{x\} </math>
+Notice the given equation is equivalent to <math>(\lfloor x \rfloor+\{x\})^2=10,000\{x\} </math>
-Now we now that <math>\{x\} < 1</math> so plugging in <math>1</math> for <math>\{x\}</math> we can find the upper and lower bounds for the values.
+Now we know that <math>\{x\} < 1</math> so plugging in <math>1</math> for <math>\{x\}</math> we can find the upper and lower bounds for the values.
 <math>(\lfloor x \rfloor +1)^2 = 10,000(1)</math>
@@ Line 103: / Line 101: @@
 ==Solution 5==
-First, we can let <math>\{x\} = b, \lfloor x \rfloor = a</math>. We know that <math>a + b = x</math> by definition. We can rearrange the equation to obtain
-<math>x^2 = 10^4(x - a)</math>.
-By taking square root on both sides, we obtain <math>x = \pm 100 \sqrt{b}</math> (because <math>x - a = b</math>). We know since <math>b</math> is the fractional part of <math>x</math>, it must be that <math>0 \leq b < 1</math>. Thus, <math>x</math> may take any value in the interval <math>-100 < x < 100</math>. Hence, we know that there are <math>\boxed{\text{(C)}~199}</math> potential values for <math>\lfloor x \rfloor</math> in that range and we are done.
-~awesome1st
-==Solution 6==
-Firstly, we can rearrange to get <math>\lfloor x \rfloor = x-\frac{x^2}{10,000}</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.
-==Solution 7==
 Firstly, if <math>x</math> is an integer, then <math>10,000\lfloor x \rfloor=10,000x</math>, so <math>x</math> must be <math>0</math>.
@@ Line 131: / Line 112: @@
 <math>0<10,000x<10,000</math>
-Therefore, <math>0<x^2+10,000\lfloor x \rfloor <1</math>, which overlaps with <math>0<10,000x<10,000</math>. This means that there is at least one real solution between <math>0</math> and <math>1</math>. Since <math>x^2+10,000\lfloor x \rfloor </math> increases exponentially and <math>10,000x</math> increases linearly, there is only one solution for this case.
+Therefore, <math>0<x^2+10,000\lfloor x \rfloor <1</math>, which overlaps with <math>0<10,000x<10,000</math>. This means that there is at least one real solution between <math>0</math> and <math>1</math>. Since <math>x^2+10,000\lfloor x \rfloor </math> increases quadratically and <math>10,000x</math> increases linearly, there is only one solution for this case.
 Similarly, if <math>1<x<2</math>, then we know the following:
@@ Line 149: / Line 130: @@
 ~Owen1204
-==Solution 8==
+==Solution 6 (General Equation)==
-General solution to this type of equation <math>f(x, \lfloor x \rfloor) = 0</math>L
+General solution to this type of equation <math>f(x, \lfloor x \rfloor) = 0</math>:
 . solve <math>f(x, \lfloor x \rfloor) = 0</math> for <math>x</math> to get <math>x = g(\lfloor x \rfloor )</math>
@@ Line 186: / Line 167: @@
 So the number of <math>k</math>'s values is <math>99-(-101)-1=199</math>. Because <math>x=5000-100\sqrt{2500-k}</math>, for each value of <math>k</math>, there is a value for <math>x</math>. The answer is <math>\boxed{\textbf{(C)} 199}</math>
-~isabelchen
+~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
+==Solution 7==
+Subtracting <math>10000\lfloor x\rfloor</math> from both sides gives <math>x^2=10000(x-\lfloor x\rfloor)=10000\{x\}</math>. Dividing both sides by <math>10000</math> gives <math>\left(\frac{x}{100}\right)^2=\{x\}<1</math>.  <math>\left(\frac{x}{100}\right)^2<1</math> when <math>-100<x<100</math> so the answer is <math>\boxed{199}</math>.
+~randomdude10807
 ==Video Solution==

2018 AMC 10B (Problems • Answer Key • Resources)
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2018 AMC 12B (Problems • Answer Key • Resources)
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