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

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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]]}}
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== Problem ==
 
== Problem ==
 
Let <math>\lfloor x \rfloor</math> denote the greatest integer less than or equal to <math>x</math>. How many real numbers <math>x</math> satisfy the equation <math>x^2 + 10,000\lfloor x \rfloor = 10,000x</math>?
 
Let <math>\lfloor x \rfloor</math> denote the greatest integer less than or equal to <math>x</math>. How many real numbers <math>x</math> satisfy the equation <math>x^2 + 10,000\lfloor x \rfloor = 10,000x</math>?
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<math>\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201</math>
 
<math>\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201</math>
  
== Solution ==
+
==Solution 1==
 
This rewrites itself to <math>x^2=10,000\{x\}</math>.
 
This rewrites itself to <math>x^2=10,000\{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.
 
 
Here is a graph of <math>y=x^2</math> and <math>y=16\{x\}</math> for visualization.
 
Here is a graph of <math>y=x^2</math> and <math>y=16\{x\}</math> for visualization.
  
Line 40: Line 41:
 
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>.
  
== Alternative, Bashy Solution ==
+
==Solution 2==
  
 
Same as the first solution, <math>x^2=10,000\{x\} </math>.
 
Same as the first solution, <math>x^2=10,000\{x\} </math>.
  
  
We can write <math>x</math> as <math>\lfloor x \rfloor+\{x\}</math>. Expanding everything, we get a quadratic in <math>{x}</math> in terms of <math>\lfloor x \rfloor</math>:
+
We can write <math>x</math> as <math>\lfloor x \rfloor+\{x\}</math>. Expanding everything, we get a quadratic in <math>\{x\}</math> in terms of <math>\lfloor x \rfloor</math>:
<math>\{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0</math>
+
<cmath> \{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0</cmath>
  
  
We use the quadratic formula to solve for {x}:
+
We use the quadratic formula to solve for <math>\{x\}</math> :
<math>\{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{( -2 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 ) }}{2} </math>
+
<cmath> \{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} </cmath>
  
  
Line 56: Line 57:
  
  
Solving over the integers, <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of<math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done.
+
Solving over the integers, <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of <math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done.
 
 
~ Alex_z_Awesome
 
  
==Another Solution==
+
==Solution 3==
  
Let <math>x = a+k</math> where <math>a</math> is the integer portion of <math>x</math> and <math>k</math> is the decimal portion.
+
Let <math>x = a+k</math> where <math>a</math> is the integer part of <math>x</math> and <math>k</math> is the fractional part of <math>x</math>.
 
We can then rewrite the problem below:
 
We can then rewrite the problem below:
  
Line 71: Line 70:
 
<math>(a+k)^2 + 10000a = 10000a + 10000k</math>
 
<math>(a+k)^2 + 10000a = 10000a + 10000k</math>
  
Solving for a+k...
+
Solving for <math>a+k = x</math>
  
 
<math>(a+k)^2 = 10000k</math>
 
<math>(a+k)^2 = 10000k</math>
  
<math>a+k = \pm100\sqrt{k}</math>
+
<math>x = a+k = \pm100\sqrt{k}</math>
 +
 
 +
Because <math>0 \leq k < 1</math>, we know that <math>a+k</math> cannot be less than or equal to <math>-100</math> nor greater than or equal to <math>100</math>. Therefore:
 +
 
 +
<math>-99 \leq x \leq 99</math>
 +
 
 +
There are <math>199</math> elements in this range, so the answer is <math>\boxed{\textbf{(C)} \text{ 199}}</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>.
 +
 
 +
==Solution 4==
 +
 
 +
Notice the given equation is equivilent 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.
 +
 
 +
<math>(\lfloor x \rfloor +1)^2 = 10,000(1)</math>
 +
 
 +
<math>(\lfloor x \rfloor +1) = \pm 100</math>
 +
 
 +
<math>\lfloor x \rfloor = 99, -101</math>
 +
 
 +
And just like <math>\textbf{Solution 2}</math>, we see that <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of <math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done.
 +
 
 +
==Solution 5==
 +
 
 +
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>.
 +
 
 +
If <math>0<x<1</math>, then we know the following:
 +
 
 +
<math>0<x^2<1</math>
 +
 
 +
<math>10,000\lfloor x \rfloor =0</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 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:
 +
 
 +
<math>1<x^2<4</math>
 +
 
 +
<math>10,000\lfloor x \rfloor =10,000</math>
 +
 
 +
<math><10,000<10,000x<20,000</math>
 +
 
 +
By following similar logic, we can find that there is one solution between <math>1</math> ad <math>2</math>.
 +
 
 +
We can also follow the same process to find that there are negative solutions for <math>x</math> as well.
 +
 
 +
There are not an infinite amount of solutions, so at one point there will be no solutions when <math>n<x<n+1</math> for some integer <math>n</math>. For there to be no solutions in a given range means that the range of <math>10,000\lfloor x \rfloor + x^2</math> does not intersect the range of <math>10,000x</math>. <math>x^2</math> will always be positive, and <math>10,000\lfloor x \rfloor</math> is less than <math>10,000</math> less than <math>10,000x</math>, so when <math>x^2 >= 10,000</math>, the equation will have no solutions. This means that there are <math>99</math> positive solutions, <math>99</math> negative solutions, and <math>0</math> for a total of <math>\boxed{\text{(C)}~199}</math> solutions.
 +
 
 +
~Owen1204
 +
 
 +
==Solution 6 (General Equation)==
 +
 
 +
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>
  
Because <math>0 <= k < 1</math>, we know that <math>a+k</math> cannot be less than or equal to <math>-100</math> nor greater than or equal to <math>100</math>. Therefore:
+
<math>x=5000 \pm 100 \sqrt{2500- \lfloor x \rfloor}</math>, <math>\lfloor x \rfloor \le 2500</math>
  
<math>-99 \leq a+k = x \leq 99</math>
+
<math>\lfloor x \rfloor \le x < \lfloor x \rfloor + 1</math>
  
There are 199 elements in this range, so the answer is <math>\fbox{C 199}</math> (MagentaCobra)
+
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>
  
edited by math4fun2
+
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
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=vHKPbaXwJUE
  
 
==See Also==
 
==See Also==
Line 91: Line 180:
 
{{AMC12 box|year=2018|ab=B|num-b=23|num-a=25}}
 
{{AMC12 box|year=2018|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Intermediate Algebra Problems]]

Latest revision as of 00:44, 6 February 2022

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 $x^2=10,000\{x\}$.

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

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 $x$.

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

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

Video Solution

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

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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 (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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