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

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~Technodoggo
 
~Technodoggo
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== Solution ==
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First, <math>x=2,1</math> are trivial solutions
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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
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We can now test values for <math>\lfloor{x}\rfloor</math>:
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<math>\lfloor{x}\rfloor=0</math>
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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
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<math>\lfloor{x}\rfloor=-1</math>
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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>
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<math>\lfloor{x}\rfloor=3</math>
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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
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<math>\lfloor{x}\rfloor=4</math>
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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.
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Our final answer is <math>\boxed{\textbf{(B) }4}</math>

Revision as of 15:59, 15 November 2023

Problem

How many distinct values of 𝑥 satisfy $\lfloor{x}\rfloor^2-3x=0.$ 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 $(\dfrac{2}{3},\dfrac{11}{3}).$

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

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

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