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) | ||
+ | |||
+ | ==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> | ||
+ | <cmath>\frac{a^2-2a+1}{9}=a-2\rightarrow a^2-2a+1=9a-18\rightarrow a^2-11a+19=0.</cmath> | ||
+ | There are no solutions in this case. | ||
+ | Case 2: <math>\lfloor x\rfloor=\frac{a-2}{3}</math> | ||
+ | <cmath>\frac{a^2-4a+4}{9}=a-2\rightarrow a^2-4a+4=9a-18\rightarrow a^2-13a+22=0.</cmath> | ||
+ | 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 21:26, 15 November 2023
Problem
How many distinct values of 𝑥 satisfy
where
denotes the largest integer less than or equal to 𝑥?
Solution (Quick)
A quadratic equation can have up to 2 real solutions. With the , 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.
, we get
as our first pair of solutions.
Up to this point, we can rule out A,E.
Next, we see that This implies that
must be an integer.
We can guess and check
as
which yields
So we got 4 in total
~Technodoggo
Solution
First, 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 :
We have . Solving, we have
. We see that
, so this solution is valid
We have . Solving, we have
.
, so this is not valid. We assume there are no more solutions in the negative direction and move on to
We have . Solving, we have
. We see that
, so this solution is valid
We have . Solving, we have
.
, so this is not valid. We assume there are no more solutions.
Our final answer is
~kjljixx
Solution
Denote .
Denote
.
Thus,
.
The equation given in this problem can be written as
Thus,
Because , we have
.
Thus,
Therefore, , 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 work.
Then, we know that
must be an integer. Set
. Now, there are two cases for the value of
.
Case 1:
There are no solutions in this case.
Case 2:
This case provides the two solutions
and
as two more solutions. Our final answer is thus
.