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Difference between revisions of "2023 AMC 10B Problems/Problem 22"
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First, we get the trivial solution by ignoring the floor.
 
First, we get the trivial solution by ignoring the floor.
−
(x-2)(x-1) = 0, we get 2,1 as solutions.
+
<math>(x-2)(x-1) = 0</math>, we get <math>(2,1)</math> as solutions.
  
−
Next, we see that <math>\lfloor{x}\rfloor^2-3x=0.</math>  This imples that -3x 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>

Revision as of 14:54, 15 November 2023

Solution

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

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 $(\dfrac{2}{3},\dfrac{11}{3}).$

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