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

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{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #25]] and [[2018 AMC 10B Problems|2018 AMC 10B #24]]}}
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{{duplicate|[[2018 AMC 10B Problems#Problem 25 | 2018 AMC 10B #25]] and [[2018 AMC 12B Problems#Problem 24|2018 AMC 12B #24]]}}
  
 
== Problem ==
 
== Problem ==
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==Solution 1==
 
==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.
 
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.
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</asy>
 
</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==
 
==Solution 2==
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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>.
 
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
 +
 +
Yet Another Note EXCEPT THIS ONE'S VERY IMPORTANT, AS IT DISPROVES THE PROOF (also not by the author of any previous notes or of the original solution): this solution claims that <math>-99 \leq x \leq 99</math>, which is not entirely true. When <math>x = 5000-1000\sqrt{26}</math>, the equation originally stated in the question holds true. <math>5000-1000\sqrt{26}</math> is roughly <math>-99.0195</math> (rounded to the nearest ten thousandth), which is not in the <math>-99 \leq x \leq 99</math> interval. It is probably by mere coincidence that this solution gives the correct answer, as the solution is not rigorous. Additionally, <math>x</math> is not necessarily integer, so it is not claimable that there are 199 elements in the range of <math>-99 \leq x \leq 99</math>.
  
 
==Solution 4==
 
==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>
 
<math>(\lfloor x \rfloor +1)^2 = 10,000(1)</math>
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==Solution 5==
 
==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>.
 
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>.
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<math>0<10,000x<10,000</math>
 
<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:
 
Similarly, if <math>1<x<2</math>, then we know the following:
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~Owen1204
 
~Owen1204
  
==Solution 8 (Quick and Dirty)==
+
==Solution 6 (General Equation)==
We can quickly realize that if x is too large, then the equation becomes invalid because x^2 grows faster than the optimal difference between the floor of x and x. We can calculate the value, and the value is 100. However, we also need to know that -100 also works. But if it is 100, it will be optimally 0.999... x 10000 which is still not 10000.  
+
 
 +
General solution to this type of equation <math>f(x, \lfloor x \rfloor) = 0</math>:
 +
 
 +
1. solve <math>f(x, \lfloor x \rfloor) = 0</math> for <math>x</math> to get <math>x = g(\lfloor x \rfloor )</math>
 +
2. apply <math>\lfloor x \rfloor \le x < \lfloor x \rfloor+1</math>, solve <math>\lfloor x \rfloor \le g(\lfloor x \rfloor) < \lfloor x \rfloor+1</math> to get the domain of <math>\lfloor x \rfloor</math>
 +
3. get <math> \lfloor x \rfloor</math> from the domain of <math> \lfloor x \rfloor</math> because <math> \lfloor x \rfloor</math> is integer, then get <math>x</math> from <math> \lfloor x \rfloor</math> by <math>x = g( \lfloor x \rfloor) </math>
 +
Note: function <math>\lfloor x \rfloor</math> maps <math>x</math> to its floor. By solving <math>f(x, \lfloor x \rfloor) = 0</math>, we get function <math>x = g( \lfloor x \rfloor) </math>, mapping <math>x</math>'s floor to <math>x</math>
 +
 
 +
<math>x^2 - 10000x + 10000 \lfloor x \rfloor =0</math>
 +
 
 +
<math>x=5000 \pm 100 \sqrt{2500- \lfloor x \rfloor}</math>, <math>\lfloor x \rfloor \le 2500</math>
 +
 
 +
<math>\lfloor x \rfloor \le x < \lfloor x \rfloor + 1</math>
 +
 
 +
If <math>x= 5000 + 100 \sqrt{2500 - \lfloor x \rfloor}</math>, <math>x \ge 5000</math>, it contradicts <math>x < \lfloor x \rfloor + 1 \le 2501</math>
 +
 
 +
So <math>x= 5000 - 100 \sqrt{2500 - \lfloor x \rfloor}</math>
 +
 
 +
Let <math>k = \lfloor x \rfloor</math> , <math>x= 5000 - 100 \sqrt{2500 - k}</math>
 +
 
 +
<math>k \le 5000 - 100 \sqrt{2500 - k} < k + 1</math>
 +
 
 +
<math>0 \le 5000 - k - 100 \sqrt{2500 - k} < 1</math>
 +
 
 +
<math>0 \le 2500 - k - 100 \sqrt{2500 - k} + 2500 < 1</math>
 +
 
 +
<math>0 \le (\sqrt{2500 - k} - 50)^2 < 1</math>
 +
 
 +
<math>-1 < \sqrt{2500 - k} - 50 < 1</math>
 +
 
 +
<math>49 < \sqrt{2500 - k} < 51</math>
 +
 
 +
<math>-101 < k < 99</math>
 +
 
 +
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>
 +
 
 +
~[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
 +
 
 +
==Solution 8 (Also Gives General Formula For Values of x)==
 +
The question wants to know how many values of <math>x</math> satisfy the equation <math>x^2 + 10,000\lfloor x \rfloor = 10,000x</math>. This equation can be simplified as follows
 +
 
 +
<cmath>x^2 + 10,000\lfloor x \rfloor = 10,000x</cmath>
 +
 
 +
<cmath>x^2 = 10,000(x-\lfloor x \rfloor)</cmath>
 +
 
 +
Notice that <math>x-\lfloor x \rfloor</math> must be greater than or equal to zero, but less than one. Because of that, the right-hand side of the equation must be less than <math>10000</math>. Therefore, <math>-100<x<100</math>.
 +
 
 +
<math>x</math> can be expressed as <math>a+b</math>, where <math>a</math> is an integer and <math>b</math> is a real number such that <math>0 \leq b < 1</math>. Notice that in order to satisfy the conditions set down for <math>x</math>, <math>-100 \leq a \leq 99</math>.
 +
 
 +
Substituting <math>a</math> and <math>b</math> into <math>x^2 = 10000(x-\lfloor x \rfloor)</math>, we get
 +
 
 +
<cmath>(a+b)^2 = 10,000b</cmath>
 +
 
 +
<cmath>a^2+2ab+b^2 = 10,000b</cmath>
 +
 
 +
Let's try to turn this into a quadratic equation where we're trying to solve for <math>b</math>. Simplifying, we find
 +
 
 +
<cmath>b^2+(2a-10,000)b+a^2 = 0</cmath>
 +
 
 +
Now the value of <math>b</math> depends on the value of <math>a</math>. Our task is to figure out how many values of <math>a</math> will give me valid values of <math>b</math> (a.k.a the values of <math>a</math> that give me <math>b</math> such that <math>0 \leq b < 1</math>), as that will be our answer. We must also keep in mind that <math>-100 \leq a \leq 99</math>.
 +
 
 +
Using the quadratic formula, we can find
 +
 
 +
<cmath>\frac{(-2a+10,000) \pm \sqrt{(2a-10,000)^2-4a^2}}{2}</cmath>
 +
 
 +
The equation above represents the values of <math>b</math> in terms of <math>a</math>. Since it represents <math>b</math>, we want values of <math>a</math> such that the equation is between zero and one, zero inclusive. We can then conceive the inequality below.
 +
 
 +
<cmath>0 \leq \frac{(-2a+10,000) \pm \sqrt{(2a-10,000)^2-4a^2}}{2} < 1</cmath>
 +
 
 +
Wow, that looks like a mess! How are we supposed to easily use such a complicated expression? Maybe it would help to simplify it further. First, let's notice a few things.
 +
 
 +
*I want the numerator to be between 0 and 2, 0 inclusive. Of the <math>\pm</math> in the quadratic formula, only the "<math>-</math>" (minus) will work. Why? Because <math>-2a+10000</math> will always be positive (if I plug in the largest possible value of <math>a</math>, <math>99</math>, it will give <math>9802</math>), and if I want <math>b</math> to be real, which I do, then <math>\sqrt{(2a-10,000)^2-4a^2}</math> should also be positive. <math>-2a+10000</math> will already be much bigger than 2, and adding more on will only make it bigger. It will never be less than 2. It will never work.
 +
 
 +
*I can simplify the numbers under the square root using Difference of Squares. I can turn <math>\sqrt{(2a-10,000)^2-4a^2}</math> into <math>\sqrt{(2a-10,000)^2-(2a)^2}</math>. Applying Difference of Squares, it will eventually simplify out to be <math>200\sqrt{2500-a}</math>.
 +
 
 +
Cool! Now we've simplified the square root and figured out that it must be a minus sign, not a plus sign. Our new inequality looks like this:
 +
 
 +
<cmath>0 \leq \frac{(-2a+10,000) - 200\sqrt{2500-a}}{2} < 1</cmath>
 +
 
 +
Simplifying the unsimplified fraction further, we can get this:
 +
 
 +
<cmath>0 \leq -a + 5,000 - 100\sqrt{2500-a} < 1</cmath>
 +
 
 +
Yay! A simpler equation. As long as the value <math>a</math> fits this inequality, the restrictions set on <math>b</math> are satisfied. But wait. What about the restrictions on <math>a</math>?
 +
 
 +
'''By testing out <math>-100</math> and <math>99</math> (and finding that <math>-100</math> gives a value very close to <math>1</math>, and that <math>99</math> gives exactly <math>1</math>), we can conclude that any value of <math>-100 \leq </math>a<math> \leq 99</math> will satisfy the inequality above as long as it does not make <math>-a + 5,000 - 100\sqrt{2500-a}</math> equal to <math>1</math>. By testing <math>-100</math> and <math>99</math>, you will also find that <math>a = 99</math> will not satisfy the inequality.''' <u><-- old, non-rigorous way. You can ignore it, but if you want to see a bogus way to prove it, feel free to read it.</u>
 +
 
 +
I can solve <math>0 \leq -a + 5000 - 100\sqrt{2500-a} < 1</math> for <math>a</math> this way:
 +
 
 +
<math>0 \leq -a + 5000 - 100\sqrt{2500-a} < 1</math>
 +
 
 +
<math>-5000 \leq -a - 100\sqrt{2500-a} < -4999</math>
 +
 
 +
<math>5000 \geq a + 100\sqrt{2500-a} < 4999</math>
 +
 
 +
<math>5000-a \geq 100\sqrt{2500-a} < 4999-a</math>
 +
 
 +
<math>25000000-10000a+a^2 \geq 25000000-10000a > 4999^2-9998a+a^2</math>
 +
 
 +
<math>a^2 \geq 0 > -9999+2a+a^2</math>
 +
 
 +
<math>0 \geq -a^2 > -9999+2a</math>
 +
 
 +
<math>0 \leq a^2 < 9999-2a</math>
 +
 
 +
<math>2a \leq a(a+2) < 9999</math>
 +
 
 +
Following the left part of the inequality,
 +
 
 +
<math>2a \leq a^2 + 2a</math>
 +
 
 +
<math>0 \leq a^2</math>
 +
 
 +
<math>a</math> can be anything <math>(-\infty, \infty)</math>.
 +
 
 +
Following the right part of the inequality,
 +
 
 +
<math>a(a+2) < 9999</math>
 +
 
 +
<math>a < 99</math> and <math>a > -101</math>
 +
 
 +
<math>a</math> can be anything <math>(-101, 99)</math>.
 +
 
 +
Combining both restrictions and given previously that <math>-100 \leq a \leq 99</math>, we can conclude that the inequality describing all possible values of <math>a</math> is <math>-100 \leq a < 99</math>.
 +
 
 +
Remember how these values of <math>a</math> give valid values of <math>b</math> that satisfy the desired inequality <math>0 \leq b < 1</math> and how adding up <math>a</math> and <math>b</math> gives <math>x</math>. In summary, remember that the number of values of <math>a</math> is equivalent to the number of values of <math>x</math>.
 +
 
 +
<math>a</math> must be integer, so the number of values of <math>a</math> that satisfy <math>-100 \leq a < 99</math> is <math>\boxed{199}</math>.
 +
 
 +
The answer is <math>\boxed{\textbf{(C)} 199}</math>.
 +
 
 +
~ :)
  
We can realize that for every Floor x , we have one x to match because we know that if x^2 is less than 100, we can adjust the value of x*10000 to match x^2. If x is larger than or equal to 10000, this is impossible. So we have -100 <math><</math> x <math>,</math> 100, so we have 199 values of x, so the answer is <math>\boxed{\text{(C)}~199}</math>.
 
  
- Arcticturn
+
Note From Author of Solution: you may have noticed that I missed a slight bit by not proving that the thing under the square root in the quadratic formula I made was positive. Don't worry, it wasn't a mistake. It's actually pretty easy to prove. I'll leave it to you to figure out how to do it!
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 11:42, 31 July 2025

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

Yet Another Note EXCEPT THIS ONE'S VERY IMPORTANT, AS IT DISPROVES THE PROOF (also not by the author of any previous notes or of the original solution): this solution claims that $-99 \leq x \leq 99$, which is not entirely true. When $x = 5000-1000\sqrt{26}$, the equation originally stated in the question holds true. $5000-1000\sqrt{26}$ is roughly $-99.0195$ (rounded to the nearest ten thousandth), which is not in the $-99 \leq x \leq 99$ interval. It is probably by mere coincidence that this solution gives the correct answer, as the solution is not rigorous. Additionally, $x$ is not necessarily integer, so it is not claimable that there are 199 elements in the range of $-99 \leq x \leq 99$.

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 $f(x, \lfloor x \rfloor) = 0$ for $x$ to get $x = g(\lfloor x \rfloor )$
2. apply $\lfloor x \rfloor \le x < \lfloor x \rfloor+1$, solve $\lfloor x \rfloor \le g(\lfloor x \rfloor) < \lfloor x \rfloor+1$ to get the domain of $\lfloor x \rfloor$
3. get $\lfloor x \rfloor$ from the domain of $\lfloor x \rfloor$ because $\lfloor x \rfloor$ is integer, then get $x$ from $\lfloor x \rfloor$ by $x = g( \lfloor x \rfloor)$
Note: function $\lfloor x \rfloor$ maps $x$ to its floor. By solving $f(x, \lfloor x \rfloor) = 0$, we get function $x = g( \lfloor x \rfloor)$, mapping $x$'s floor to $x$

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

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

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

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

Solution 8 (Also Gives General Formula For Values of x)

The question wants to know how many values of $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$. This equation can be simplified as follows

\[x^2 + 10,000\lfloor x \rfloor = 10,000x\]

\[x^2 = 10,000(x-\lfloor x \rfloor)\]

Notice that $x-\lfloor x \rfloor$ must be greater than or equal to zero, but less than one. Because of that, the right-hand side of the equation must be less than $10000$. Therefore, $-100<x<100$.

$x$ can be expressed as $a+b$, where $a$ is an integer and $b$ is a real number such that $0 \leq b < 1$. Notice that in order to satisfy the conditions set down for $x$, $-100 \leq a \leq 99$.

Substituting $a$ and $b$ into $x^2 = 10000(x-\lfloor x \rfloor)$, we get

\[(a+b)^2 = 10,000b\]

\[a^2+2ab+b^2 = 10,000b\]

Let's try to turn this into a quadratic equation where we're trying to solve for $b$. Simplifying, we find

\[b^2+(2a-10,000)b+a^2 = 0\]

Now the value of $b$ depends on the value of $a$. Our task is to figure out how many values of $a$ will give me valid values of $b$ (a.k.a the values of $a$ that give me $b$ such that $0 \leq b < 1$), as that will be our answer. We must also keep in mind that $-100 \leq a \leq 99$.

Using the quadratic formula, we can find

\[\frac{(-2a+10,000) \pm \sqrt{(2a-10,000)^2-4a^2}}{2}\]

The equation above represents the values of $b$ in terms of $a$. Since it represents $b$, we want values of $a$ such that the equation is between zero and one, zero inclusive. We can then conceive the inequality below.

\[0 \leq \frac{(-2a+10,000) \pm \sqrt{(2a-10,000)^2-4a^2}}{2} < 1\]

Wow, that looks like a mess! How are we supposed to easily use such a complicated expression? Maybe it would help to simplify it further. First, let's notice a few things.

  • I want the numerator to be between 0 and 2, 0 inclusive. Of the $\pm$ in the quadratic formula, only the "$-$" (minus) will work. Why? Because $-2a+10000$ will always be positive (if I plug in the largest possible value of $a$, $99$, it will give $9802$), and if I want $b$ to be real, which I do, then $\sqrt{(2a-10,000)^2-4a^2}$ should also be positive. $-2a+10000$ will already be much bigger than 2, and adding more on will only make it bigger. It will never be less than 2. It will never work.
  • I can simplify the numbers under the square root using Difference of Squares. I can turn $\sqrt{(2a-10,000)^2-4a^2}$ into $\sqrt{(2a-10,000)^2-(2a)^2}$. Applying Difference of Squares, it will eventually simplify out to be $200\sqrt{2500-a}$.

Cool! Now we've simplified the square root and figured out that it must be a minus sign, not a plus sign. Our new inequality looks like this:

\[0 \leq \frac{(-2a+10,000) - 200\sqrt{2500-a}}{2} < 1\]

Simplifying the unsimplified fraction further, we can get this:

\[0 \leq -a + 5,000 - 100\sqrt{2500-a} < 1\]

Yay! A simpler equation. As long as the value $a$ fits this inequality, the restrictions set on $b$ are satisfied. But wait. What about the restrictions on $a$?

By testing out $-100$ and $99$ (and finding that $-100$ gives a value very close to $1$, and that $99$ gives exactly $1$), we can conclude that any value of $-100 \leq$a$\leq 99$ will satisfy the inequality above as long as it does not make $-a + 5,000 - 100\sqrt{2500-a}$ equal to $1$. By testing $-100$ and $99$, you will also find that $a = 99$ will not satisfy the inequality. <-- old, non-rigorous way. You can ignore it, but if you want to see a bogus way to prove it, feel free to read it.

I can solve $0 \leq -a + 5000 - 100\sqrt{2500-a} < 1$ for $a$ this way:

$0 \leq -a + 5000 - 100\sqrt{2500-a} < 1$

$-5000 \leq -a - 100\sqrt{2500-a} < -4999$

$5000 \geq a + 100\sqrt{2500-a} < 4999$

$5000-a \geq 100\sqrt{2500-a} < 4999-a$

$25000000-10000a+a^2 \geq 25000000-10000a > 4999^2-9998a+a^2$

$a^2 \geq 0 > -9999+2a+a^2$

$0 \geq -a^2 > -9999+2a$

$0 \leq a^2 < 9999-2a$

$2a \leq a(a+2) < 9999$

Following the left part of the inequality,

$2a \leq a^2 + 2a$

$0 \leq a^2$

$a$ can be anything $(-\infty, \infty)$.

Following the right part of the inequality,

$a(a+2) < 9999$

$a < 99$ and $a > -101$

$a$ can be anything $(-101, 99)$.

Combining both restrictions and given previously that $-100 \leq a \leq 99$, we can conclude that the inequality describing all possible values of $a$ is $-100 \leq a < 99$.

Remember how these values of $a$ give valid values of $b$ that satisfy the desired inequality $0 \leq b < 1$ and how adding up $a$ and $b$ gives $x$. In summary, remember that the number of values of $a$ is equivalent to the number of values of $x$.

$a$ must be integer, so the number of values of $a$ that satisfy $-100 \leq a < 99$ is $\boxed{199}$.

The answer is $\boxed{\textbf{(C)} 199}$.

~ :)


Note From Author of Solution: you may have noticed that I missed a slight bit by not proving that the thing under the square root in the quadratic formula I made was positive. Don't worry, it wasn't a mistake. It's actually pretty easy to prove. I'll leave it to you to figure out how to do it!

Video Solution

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

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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Problem 24
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Last Problem
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All AMC 10 Problems and Solutions
2018 AMC 12B (ProblemsAnswer KeyResources)
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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