Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 10B Problems/Problem 22"

(→‎Solution)
(→‎Similar approach as Solution 8)
 
(80 intermediate revisions by 24 users not shown)
Line 1: Line 1:
−
== Solution ==
+
== Problem ==
 +
 
 +
How many distinct values of <math>x</math> satisfy
 +
<math>\lfloor{x}\rfloor^2-3x+2=0</math>, where <math>\lfloor{x}\rfloor</math> denotes the largest integer less than or equal to <math>x</math>?
 +
 
 +
<math>\textbf{(A) } \text{an infinite number} \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 0</math>
 +
 
 +
==Solution 1==
 +
 
 +
To further grasp at this equation, we rearrange the equation into
 +
<cmath>\lfloor{x}\rfloor^2=3x-2.</cmath>
 +
Thus, <math>3x-2</math> is a perfect square and nonnegative. It is now much more apparent that <math>x \ge 2/3,</math> and that <math>x = 2/3</math> is a solution.
 +
 
 +
Additionally, by observing the RHS, <math>x<4,</math> as
 +
<cmath>\lfloor{4}\rfloor^2 > 3\cdot4,</cmath>
 +
since squares grow quicker than linear functions.
 +
 
 +
Now that we have narrowed down our search, we can simply test for intervals <math>[2/3,1], [1,2],[2,3],[3,4).</math> This intuition to use intervals stems from the fact that <math>x=1,2</math> are observable integral solutions.
 +
 
 +
Notice how there is only one solution per interval, as <math>3x-2</math> increases while <math>\lfloor{x}\rfloor^2</math> stays the same.
 +
 
 +
Finally, we see that <math>x=3</math> does not work, however, through setting <math>\lfloor{x}\rfloor^2 = 9,</math> <math>x = 11/3</math> is a solution and within our domain of <math>[3,4).</math>
 +
 
 +
This provides us with solutions <math>\left(\frac23, 1, 2, \frac{11}{3}\right),</math> thus the final answer is <math>\boxed{(\text{B}) \ 4}.</math>
 +
 
 +
~mathbrek, happyhari
 +
 
 +
==Solution 2 (Desperation)==
 +
Notice there has to be a solution for <math>x</math> between <math>(2,-3)</math> and <math>(1,2)</math> because of the floors. There is also no way <math>2</math> solutions because of the quadratic, and when we add them together, we get <math>\boxed{(\text{B}) \ 4}.</math>
 +
~perion.
 +
 
 +
==Solution 3 (Three Cases)==
 +
First, let's take care of the integer case--clearly, only <math>x=1,2</math> work.
 +
Then, we know that <math>3x</math> must be an integer. Set <math>x=\frac{a}3</math>. Now, there are two cases for the value of <math>\lfloor x\rfloor</math>.
 +
Case 1: <math>\lfloor x\rfloor=\frac{a-1}{3}</math>
 +
<cmath>\frac{a^2-2a+1}{9}=a-2\rightarrow a^2-2a+1=9a-18\rightarrow a^2-11a+19=0.</cmath>
 +
There are no solutions in this case.
 +
Case 2: <math>\lfloor x\rfloor=\frac{a-2}{3}</math>
 +
<cmath>\frac{a^2-4a+4}{9}=a-2\rightarrow a^2-4a+4=9a-18\rightarrow a^2-13a+22=0.</cmath>
 +
This case provides the two solutions <math>\frac23</math> and <math>\frac{11}3</math> as two more solutions. Our final answer is thus <math>\boxed{4}</math>.
 +
 
 +
~wuwang2002
 +
 
 +
== Solution 4==
 +
 
 +
First, <math>x=2,1</math> are trivial solutions
 +
 
 +
We assume from the shape of a parabola and the nature of the floor function that any additional roots will be near 2 and 1
 +
 
 +
We can now test values for <math>\lfloor{x}\rfloor</math>:
 +
 
 +
<math>\lfloor{x}\rfloor=0</math>
 +
 
 +
We have <math>0-3x+2=0</math>. Solving, we have <math>x=\frac{2}{3}</math>. We see that <math>\lfloor{\frac{2}{3}}\rfloor=0</math>, so this solution is valid
 +
 
 +
<math>\lfloor{x}\rfloor=-1</math>
 +
 
 +
We have <math>1-3x+2=0</math>. Solving, we have <math>x=1</math>. <math>\lfloor{1}\rfloor\neq-1</math>, so this is not valid. We assume there are no more solutions in the negative direction and move on to <math>\lfloor{x}\rfloor=3</math>
 +
 
 +
<math>\lfloor{x}\rfloor=3</math>
 +
 
 +
We have <math>9-3x+2=0</math>. Solving, we have <math>x=\frac{11}{3}</math>. We see that <math>\lfloor{\frac{11}{3}}\rfloor=3</math>, so this solution is valid
 +
 
 +
<math>\lfloor{x}\rfloor=4</math>
 +
 
 +
We have <math>16-3x+2=0</math>. Solving, we have <math>x=6</math>. <math>\lfloor{6}\rfloor\neq4</math>, so this is not valid. We assume there are no more solutions.
 +
 
 +
Our final answer is <math>\boxed{\textbf{(B) }4}</math>
 +
 
 +
~kjljixx
 +
 
 +
==Solution 5==
 +
 
 +
Denote <math>a = \lfloor x \rfloor</math>.
 +
Denote <math>b = x - \lfloor x \rfloor</math>.
 +
Thus, <math>b \in \left[ 0 , 1 \right)</math>.
 +
 
 +
The equation given in this problem can be written as
 +
<cmath>
 +
\[
 +
a^2 - 3 \left( a + b \right) + 2 = 0 .
 +
\]
 +
</cmath>
 +
 
 +
Thus,
 +
<cmath>
 +
\begin{align*}
 +
3 b & = a^2 - 3 a + 2 .
 +
\end{align*}
 +
</cmath>
 +
 
 +
Because <math>b \in \left[ 0 , 1 \right)</math>, we have <math>3 b \in \left[ 0 , 3 \right)</math>.
 +
Thus,
 +
<cmath>
 +
\[
 +
a^2 - 3 a + 2 = 0, 1, \mbox{ or } 2 .
 +
\]
 +
</cmath>
 +
 
 +
If <math>a^2-3a+2=0</math>, <math>(a-2)(a-1)=0</math> so <math>a</math> can be <math>1, 2</math>.
 +
 
 +
If <math>a^2-3a+2=1</math>, <math>a^2-3a+1=0</math> which we find has no integer solutions after finding the discriminant.
 +
 
 +
If <math>a^2-3a+2=2</math>, <math>a^2-3a=0</math> -> <math>a(a-3)=0</math> so <math>a</math> can also be <math>0, 3</math>.
 +
 
 +
Therefore, <math>a = 1</math>, 2, 0, 3.
 +
Therefore, the number of solutions is
 +
<math>\boxed{\textbf{(B) 4}}</math>.
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
== Solution 6 (Quick) ==
 +
 
 +
A quadratic equation can have up to 2 real solutions. With the <math>\lfloor{x}\rfloor</math>, it could also help generate another pair. We have to verify that the solutions are real and distinct.
 +
 
  
 
First, we get the trivial solution by ignoring the floor.
 
First, we get the trivial solution by ignoring the floor.
−
<math>(x-2)(x-1) = 0</math>, we get <math>(2,1)</math> as solutions.
+
<math>(x-2)(x-1) = 0</math>, we get <math>(2,1)</math> as our first pair of solutions.
 +
 
 +
Up to this point, we can rule out A,E.
  
 
Next, we see that <math>\lfloor{x}\rfloor^2-3x=0.</math>  This implies that <math>-3x</math> must be an integer.
 
Next, we see that <math>\lfloor{x}\rfloor^2-3x=0.</math>  This implies that <math>-3x</math> must be an integer.
−
We can guess and check <math>x</math> as <math>\dfrac{k}{3}</math> which yields <math>(\dfrac{2}{3},\dfrac{11}{3}).</math>
+
We can guess and check <math>x</math> as <math>\dfrac{k}{3}</math> which yields <math>\left(\dfrac{2}{3},\dfrac{11}{3}\right).</math>
 +
 
 +
So we got 4 in total <math>\left(\dfrac{2}{3},1,2,\dfrac{11}{3}\right).</math>
 +
 
 +
~Technodoggo
 +
 
 +
== Solution 7 ==
 +
<math>x=1, 2</math> are trivial solutions.
 +
Let <math>x=n+f</math> for some integer <math>n</math> and some number <math>f</math> such that <math>-1<f<1</math>. <cmath>\lfloor{x}\rfloor^2-3x+2= \lfloor{n+f}\rfloor^2-3(n+f)+2=n^2+-3(n+f)+2.</cmath> So now we have
 +
<cmath>n^2-3(n+f)+2 = 0,</cmath>
 +
which we can rewrite as
 +
<cmath>n(n-3)=3f-2.</cmath>
 +
Since <math>n</math> is an integer, <math>n(n-3)</math> is an integer, so <math>3f-2</math> is an integer. Since <math>-1<f<1</math>, the only possible values of <math>f</math> are <math>\frac{1}{3}</math>, <math>\frac{2}{3}</math>, <math>-\frac{1}{3}</math>, and <math>-\frac{2}{3}</math>. Plugging in each value, we find that the only value of <math>f</math> that produces integer solutions for <math>n</math> is <math>f=\frac{2}{3}</math>. If <math>f=\frac{2}{3}</math>, <math>n=0</math> or <math>n=3</math>. Hence, there is a total of 4 possible solutions, so the answer is <math>\boxed{\textbf{(B) }4}</math>.
 +
~azc1027
 +
 
 +
==Solution 8==
 +
 
 +
We rewrite the equation as <math>{\lfloor x\rfloor}^2-3\lfloor x\rfloor-3\{x\}+2=0</math>, where <math>\{x\}</math> is the fractional part of <math>x</math>
 +
 
 +
Denote <math>\lfloor x\rfloor = x_1</math> and <math>\{x\} = x_2.</math> Thus
 +
<cmath>{x_1}^2-3{x_1}-3{x_2}+2=0.</cmath>
 +
 
 +
By definition, <math>0\leq x_2\leq 1</math>. We then have <math>{x_1}^2-3{x_1}+2=3{x_2}</math> and therefore <math>0\leq {x_1}^2-3{x_1}+2\leq 3</math>.
 +
 
 +
Solving, we have <math>
 +
\left[\frac{3-\sqrt{13}}{2},1\right]\cup \left[2,\frac{3+\sqrt{13}}{2}\right]</math>. But since <math>x_1</math> is an integer, we have <math>x_1</math> can only be <math>0,1,2,</math> or <math>3</math>.
 +
 
 +
Testing, we see these values of <math>x_1</math> work, and therefore the answer is just <math>\boxed{\textbf{(B) }4}</math>.
 +
 
 +
~ESAOPS
 +
 
 +
==Similar approach as Solution 8==
 +
Use the fact that <math>x = \lfloor x \rfloor + \{x\}</math>. Thus we have
 +
<cmath>(\lfloor x \rfloor^2 - 3\lfloor x \rfloor + 2) - 3\{x\} = 0.</cmath>
 +
 
 +
Noting that <math>0 \leq \{x\} < 1</math>, we get
 +
 
 +
<cmath>0 \leq (\lfloor x \rfloor - 2)(\lfloor x \rfloor - 1) < 3.</cmath>
 +
 
 +
From there, it is not too hard to see that the only values of <math>\lfloor x \rfloor</math> that satisfy this condition (while also noting that <math>\lfloor x \rfloor</math> must be an integer) are 3, 2, 1, and 0, yielding 4 values.
 +
 
 +
~mathboy282
 +
 
 +
==Solution 9 (Very Fast)==
 +
We know that for integer values of x, the graph is just <math>x^2-3x+2</math>. From the interval <math>[x, x+1]</math>, the square stays the same, so the graph has a line segment that goes down by 3 and right by 2. This is very easy to graph, so we see that there are 4 solutions. Or, we notice that only <math>x = 0, 1, 2, 3</math> results in a <math>x^2-3x+2</math> in the interval <math>[0, 3]</math>.That is <math>\boxed{\textbf{(B) }4}</math> solutions.
 +
 
 +
~Xyco
 +
 
 +
==Solution 10(has 3  cases) ==
 +
 
 +
define <math>[x] = n</math>
 +
 
 +
define the fractional part of <math>x</math> as <math>x(f)</math>
 +
 
 +
thus <math>\lfloor{x}\rfloor^2-3x+2=0</math> is
 +
 
 +
<math>n^2-3(n+x(f))+2=0</math>
 +
 
 +
<math>n^2-3n-3x(f)+2=0</math>
 +
 
 +
<math>n^2-3n+2</math> must always be an integer
 +
and thus for this equal zero
 +
<math>3x(f)</math> must also equal integer
 +
 
 +
thus <math>x(f)</math> must be fraction <math>q/3</math>
 +
and q must be 0,1,2
 +
 
 +
thus <math>x(f)</math> must be 1/3 or 2/3 or 0/3
 +
 
 +
plugging in all
 +
 
 +
<math>n^2-3n-3(0)+2=0</math>
 +
 
 +
<math>n^2-3n-3(2/3)+2=0</math>
 +
 
 +
<math>n^2-3n-3(1/3)+2=0</math>
 +
 
 +
we simplify into
 +
 +
<math>n^2-3n+2=0</math>
 +
 
 +
<math>n^2-3n=0</math>
 +
 
 +
<math>n^2-3n+1</math>
 +
 
 +
where n must be a integer
 +
 
 +
just use solve for n and use only integers
 +
 
 +
we get 2 integers for the first for <math>x(f)=0</math>
 +
 
 +
<math>n=1,2</math>
 +
 
 +
<math>n+x(f)=x</math>
 +
 
 +
<math>1+0=1</math>
 +
 
 +
<math>2+0=2</math>
 +
 
 +
<math>x=1,2</math>
 +
 
 +
we get 2 integers for the second for <math>x(f)=2/3</math>
 +
 
 +
<math>n=0,3</math>
 +
 
 +
<math>n+x(f)=x</math>
 +
 
 +
<math>0+2/3=2/3</math>
 +
 
 +
<math>3+2/3=11/3</math>
 +
 
 +
We get ZERO integers for the third for <math>x(f)=1/3</math>
 +
 
 +
we get use the quadratic discriminant to see
 +
<math>\sqrt{b^{2}-4ac}</math>
 +
 
 +
Our equation is <math>n^2-3n+1</math>
 +
 
 +
<math>\sqrt{3^{2}-4(1)(1)}</math>
 +
yielding a non integer value which means this case is invalid
 +
 
 +
 
 +
we count a total of 4 solutions which are <math>x=2/3,1,2,11/3</math>
 +
 
 +
Our answer is
 +
<math>\boxed{\textbf{(B) }4}</math>
 +
 
 +
==Solution 11 (Based on graph)==
 +
 
 +
[[File:2023AMC10BQ22Solution.jpg|400px]]
 +
 
 +
For <math>\lfloor x \rfloor^2 - 3x + 2 = 0</math>, there is a discontinuity at each integer value of <math>x</math>, and it also lies on the non-floor version of the function. Between each integer <math>x</math> and the next forms a line with a slope of <math>-3</math>. This simplifies the task of sketching the function's graph. Note that the points at <math>1</math> and <math>2</math> are considered intersections because they are points on the left side of each integer interval that exactly lie on <math>y = 0</math>. Thus, we conclude there are <math>4</math> intersection points, and the answer is <math>\boxed{(\text{B}) \ 4}</math>.
 +
～Athmyx
 +
 
 +
==Video Solution 1 by OmegaLearn==
 +
https://youtu.be/wAYcpn-Q_KQ
 +
 
 +
==Video Solution 2 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=DvHGEXBjf0Y
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/ONRoop23LIY
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
==See also==
 +
{{AMC10 box|year=2023|ab=B|num-b=21|num-a=23}}
 +
{{MAA Notice}}

Latest revision as of 21:50, 25 September 2024

Problem

How many distinct values of $x$ satisfy $\lfloor{x}\rfloor^2-3x+2=0$, where $\lfloor{x}\rfloor$ denotes the largest integer less than or equal to $x$?

$\textbf{(A) } \text{an infinite number} \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 0$

Solution 1

To further grasp at this equation, we rearrange the equation into \[\lfloor{x}\rfloor^2=3x-2.\] Thus, $3x-2$ is a perfect square and nonnegative. It is now much more apparent that $x \ge 2/3,$ and that $x = 2/3$ is a solution.

Additionally, by observing the RHS, $x<4,$ as \[\lfloor{4}\rfloor^2 > 3\cdot4,\] since squares grow quicker than linear functions.

Now that we have narrowed down our search, we can simply test for intervals $[2/3,1], [1,2],[2,3],[3,4).$ This intuition to use intervals stems from the fact that $x=1,2$ are observable integral solutions.

Notice how there is only one solution per interval, as $3x-2$ increases while $\lfloor{x}\rfloor^2$ stays the same.

Finally, we see that $x=3$ does not work, however, through setting $\lfloor{x}\rfloor^2 = 9,$ $x = 11/3$ is a solution and within our domain of $[3,4).$

This provides us with solutions $\left(\frac23, 1, 2, \frac{11}{3}\right),$ thus the final answer is $\boxed{(\text{B}) \ 4}.$

~mathbrek, happyhari

Solution 2 (Desperation)

Notice there has to be a solution for $x$ between $(2,-3)$ and $(1,2)$ because of the floors. There is also no way $2$ solutions because of the quadratic, and when we add them together, we get $\boxed{(\text{B}) \ 4}.$ ~perion.

Solution 3 (Three Cases)

First, let's take care of the integer case--clearly, only $x=1,2$ work. Then, we know that $3x$ must be an integer. Set $x=\frac{a}3$. Now, there are two cases for the value of $\lfloor x\rfloor$. Case 1: $\lfloor x\rfloor=\frac{a-1}{3}$ \[\frac{a^2-2a+1}{9}=a-2\rightarrow a^2-2a+1=9a-18\rightarrow a^2-11a+19=0.\] There are no solutions in this case. Case 2: $\lfloor x\rfloor=\frac{a-2}{3}$ \[\frac{a^2-4a+4}{9}=a-2\rightarrow a^2-4a+4=9a-18\rightarrow a^2-13a+22=0.\] This case provides the two solutions $\frac23$ and $\frac{11}3$ as two more solutions. Our final answer is thus $\boxed{4}$.

~wuwang2002

Solution 4

First, $x=2,1$ are trivial solutions

We assume from the shape of a parabola and the nature of the floor function that any additional roots will be near 2 and 1

We can now test values for $\lfloor{x}\rfloor$:

$\lfloor{x}\rfloor=0$

We have $0-3x+2=0$. Solving, we have $x=\frac{2}{3}$. We see that $\lfloor{\frac{2}{3}}\rfloor=0$, so this solution is valid

$\lfloor{x}\rfloor=-1$

We have $1-3x+2=0$. Solving, we have $x=1$. $\lfloor{1}\rfloor\neq-1$, so this is not valid. We assume there are no more solutions in the negative direction and move on to $\lfloor{x}\rfloor=3$

$\lfloor{x}\rfloor=3$

We have $9-3x+2=0$. Solving, we have $x=\frac{11}{3}$. We see that $\lfloor{\frac{11}{3}}\rfloor=3$, so this solution is valid

$\lfloor{x}\rfloor=4$

We have $16-3x+2=0$. Solving, we have $x=6$. $\lfloor{6}\rfloor\neq4$, so this is not valid. We assume there are no more solutions.

Our final answer is $\boxed{\textbf{(B) }4}$

~kjljixx

Solution 5

Denote $a = \lfloor x \rfloor$. Denote $b = x - \lfloor x \rfloor$. Thus, $b \in \left[ 0 , 1 \right)$.

The equation given in this problem can be written as \[ a^2 - 3 \left( a + b \right) + 2 = 0 . \]

Thus, \begin{align*} 3 b & = a^2 - 3 a + 2 . \end{align*}

Because $b \in \left[ 0 , 1 \right)$, we have $3 b \in \left[ 0 , 3 \right)$. Thus, \[ a^2 - 3 a + 2 = 0, 1, \mbox{ or } 2 . \]

If $a^2-3a+2=0$, $(a-2)(a-1)=0$ so $a$ can be $1, 2$.

If $a^2-3a+2=1$, $a^2-3a+1=0$ which we find has no integer solutions after finding the discriminant.

If $a^2-3a+2=2$, $a^2-3a=0$ -> $a(a-3)=0$ so $a$ can also be $0, 3$.

Therefore, $a = 1$, 2, 0, 3. Therefore, the number of solutions is $\boxed{\textbf{(B) 4}}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 6 (Quick)

A quadratic equation can have up to 2 real solutions. With the $\lfloor{x}\rfloor$, it could also help generate another pair. We have to verify that the solutions are real and distinct.


First, we get the trivial solution by ignoring the floor. $(x-2)(x-1) = 0$, we get $(2,1)$ as our first pair of solutions.

Up to this point, we can rule out A,E.

Next, we see that $\lfloor{x}\rfloor^2-3x=0.$ This implies that $-3x$ must be an integer. We can guess and check $x$ as $\dfrac{k}{3}$ which yields $\left(\dfrac{2}{3},\dfrac{11}{3}\right).$

So we got 4 in total $\left(\dfrac{2}{3},1,2,\dfrac{11}{3}\right).$

~Technodoggo

Solution 7

$x=1, 2$ are trivial solutions. Let $x=n+f$ for some integer $n$ and some number $f$ such that $-1<f<1$. \[\lfloor{x}\rfloor^2-3x+2= \lfloor{n+f}\rfloor^2-3(n+f)+2=n^2+-3(n+f)+2.\] So now we have \[n^2-3(n+f)+2 = 0,\] which we can rewrite as \[n(n-3)=3f-2.\] Since $n$ is an integer, $n(n-3)$ is an integer, so $3f-2$ is an integer. Since $-1<f<1$, the only possible values of $f$ are $\frac{1}{3}$, $\frac{2}{3}$, $-\frac{1}{3}$, and $-\frac{2}{3}$. Plugging in each value, we find that the only value of $f$ that produces integer solutions for $n$ is $f=\frac{2}{3}$. If $f=\frac{2}{3}$, $n=0$ or $n=3$. Hence, there is a total of 4 possible solutions, so the answer is $\boxed{\textbf{(B) }4}$. ~azc1027

Solution 8

We rewrite the equation as ${\lfloor x\rfloor}^2-3\lfloor x\rfloor-3\{x\}+2=0$, where $\{x\}$ is the fractional part of $x$

Denote $\lfloor x\rfloor = x_1$ and $\{x\} = x_2.$ Thus \[{x_1}^2-3{x_1}-3{x_2}+2=0.\]

By definition, $0\leq x_2\leq 1$. We then have ${x_1}^2-3{x_1}+2=3{x_2}$ and therefore $0\leq {x_1}^2-3{x_1}+2\leq 3$.

Solving, we have $\left[\frac{3-\sqrt{13}}{2},1\right]\cup \left[2,\frac{3+\sqrt{13}}{2}\right]$. But since $x_1$ is an integer, we have $x_1$ can only be $0,1,2,$ or $3$.

Testing, we see these values of $x_1$ work, and therefore the answer is just $\boxed{\textbf{(B) }4}$.

~ESAOPS

Similar approach as Solution 8

Use the fact that $x = \lfloor x \rfloor + \{x\}$. Thus we have \[(\lfloor x \rfloor^2 - 3\lfloor x \rfloor + 2) - 3\{x\} = 0.\]

Noting that $0 \leq \{x\} < 1$, we get

\[0 \leq (\lfloor x \rfloor - 2)(\lfloor x \rfloor - 1) < 3.\]

From there, it is not too hard to see that the only values of $\lfloor x \rfloor$ that satisfy this condition (while also noting that $\lfloor x \rfloor$ must be an integer) are 3, 2, 1, and 0, yielding 4 values.

~mathboy282

Solution 9 (Very Fast)

We know that for integer values of x, the graph is just $x^2-3x+2$. From the interval $[x, x+1]$, the square stays the same, so the graph has a line segment that goes down by 3 and right by 2. This is very easy to graph, so we see that there are 4 solutions. Or, we notice that only $x = 0, 1, 2, 3$ results in a $x^2-3x+2$ in the interval $[0, 3]$.That is $\boxed{\textbf{(B) }4}$ solutions.

~Xyco

Solution 10(has 3 cases)

define $[x] = n$

define the fractional part of $x$ as $x(f)$

thus $\lfloor{x}\rfloor^2-3x+2=0$ is

$n^2-3(n+x(f))+2=0$

$n^2-3n-3x(f)+2=0$

$n^2-3n+2$ must always be an integer and thus for this equal zero $3x(f)$ must also equal integer

thus $x(f)$ must be fraction $q/3$ 

and q must be 0,1,2

thus $x(f)$ must be 1/3 or 2/3 or 0/3

plugging in all

$n^2-3n-3(0)+2=0$

$n^2-3n-3(2/3)+2=0$

$n^2-3n-3(1/3)+2=0$

we simplify into

$n^2-3n+2=0$

$n^2-3n=0$

$n^2-3n+1$

where n must be a integer

just use solve for n and use only integers

we get 2 integers for the first for $x(f)=0$

$n=1,2$

$n+x(f)=x$

$1+0=1$

$2+0=2$

$x=1,2$

we get 2 integers for the second for $x(f)=2/3$

$n=0,3$

$n+x(f)=x$

$0+2/3=2/3$

$3+2/3=11/3$

We get ZERO integers for the third for $x(f)=1/3$

we get use the quadratic discriminant to see $\sqrt{b^{2}-4ac}$

Our equation is $n^2-3n+1$

$\sqrt{3^{2}-4(1)(1)}$ yielding a non integer value which means this case is invalid


we count a total of 4 solutions which are $x=2/3,1,2,11/3$

Our answer is 

$\boxed{\textbf{(B) }4}$

Solution 11 (Based on graph)

2023AMC10BQ22Solution.jpg

For $\lfloor x \rfloor^2 - 3x + 2 = 0$, there is a discontinuity at each integer value of $x$, and it also lies on the non-floor version of the function. Between each integer $x$ and the next forms a line with a slope of $-3$. This simplifies the task of sketching the function's graph. Note that the points at $1$ and $2$ are considered intersections because they are points on the left side of each integer interval that exactly lie on $y = 0$. Thus, we conclude there are $4$ intersection points, and the answer is $\boxed{(\text{B}) \ 4}$. ～Athmyx

Video Solution 1 by OmegaLearn

https://youtu.be/wAYcpn-Q_KQ

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=DvHGEXBjf0Y

Video Solution

https://youtu.be/ONRoop23LIY

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10B_Problems/Problem_22&oldid=228043"