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

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== Solution ==
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== Solution (Quick) ==
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A quadric equations 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.
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First, we get the trivial solution by ignoring the floor.
 
First, we get the trivial solution by ignoring the floor.
<math>(x-2)(x-1) = 0</math>, we get <math>(2,1)</math> as solutions.
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<math>(x-2)(x-1) = 0</math>, we get <math>(2,1)</math> as our first pair of solutions.
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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.
 
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>
 
We can guess and check <math>x</math> as <math>\dfrac{k}{3}</math> which yields <math>(\dfrac{2}{3},\dfrac{11}{3}).</math>
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So we got 4 in total <math>(\dfrac{2}{3},1,2,\dfrac{11}{3}).</math>
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~Technodoggo

Revision as of 14:02, 15 November 2023

Solution (Quick)

A quadric equations 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