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

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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Solution (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>
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<cmath>\frac{a^2-2a+1}{9}=a-2\rightarrow a^2-2a+1=9a-18\rightarrow a^2-11a+19=0.</cmath>
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There are no solutions in this case.
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Case 2: <math>\lfloor x\rfloor=\frac{a-2}{3}</math>
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<cmath>\frac{a^2-4a+4}{9}=a-2\rightarrow a^2-4a+4=9a-18\rightarrow a^2-13a+22=0.</cmath>
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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>.

Revision as of 20:26, 15 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?

Solution (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

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

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 (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 $\boxed{4}$.

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