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

Revision as of 15:35, 16 November 2023

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(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 2

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 3

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, $a^2 - 3 a + 2 = 0, 1, \mbox{ or } 2 .$

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 4(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

Video Solution 1 by OmegaLearn

https://youtu.be/wAYcpn-Q_KQ

Video Solution 2 by SpreadTheMathLove

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

@@ Line 6: / Line 6: @@
 <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 (Quick) ==
+==Solution 1(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>.
-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.
+~wuwang2002
-First, we get the trivial solution by ignoring the floor.
-<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.
-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 ==
+== Solution 2==
 First, <math>x=2,1</math> are trivial solutions
@@ Line 51: / Line 46: @@
 ~kjljixx
-==Solution==
+==Solution 3==
 Denote <math>a = \lfloor x \rfloor</math>.
@@ Line 85: / Line 80: @@
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
-==Solution (three cases)==
+== Solution 4(Quick) ==
-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>.
+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.
-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.
+First, we get the trivial solution by ignoring the floor.
-Case 2: <math>\lfloor x\rfloor=\frac{a-2}{3}</math>
+<math>(x-2)(x-1) = 0</math>, we get <math>(2,1)</math> as our first pair of solutions.
-<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>.
+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.
+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>
-~wuwang2002
+~Technodoggo
 ==Video Solution 1 by OmegaLearn==

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