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

Revision as of 11:51, 20 April 2024

1 Problem
2 Solution 1
3 Solution 2 (Desperate)
4 Solution 3 (Three Cases)
5 Solution 4
6 Solution 5
7 Solution 6 (Quick)
8 Solution 7
9 Solution 8
10 Solution 9 (Very Fast)
11 Solution 10(has 3 cases)
12 Solution 11 (Based on graph)
13 Video Solution 1 by OmegaLearn
14 Video Solution 2 by SpreadTheMathLove
15 Video Solution
16 See also

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 (Desperate)

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

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

Solution 11 (Based on graph)

Here is the translated sentence with $x$ replaced by $x$ in LaTeX format:

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 $\boxed{(\text{B}) \ 4}$.

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