Difference between revisions of "2023 AMC 10B Problems/Problem 22"
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== Problem == | == Problem == | ||
− | How many distinct values of | + | How many distinct values of <math>x</math> satisfy |
<math>\lfloor{x}\rfloor^2-3x+2=0</math>, where <math>\lfloor{x}\rfloor</math> denotes the largest integer less than or equal to <math>x</math>? | <math>\lfloor{x}\rfloor^2-3x+2=0</math>, where <math>\lfloor{x}\rfloor</math> denotes the largest integer less than or equal to <math>x</math>? | ||
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~mathbrek, happyhari | ~mathbrek, happyhari | ||
− | + | ==Solution 2 (Desperation)== | |
− | + | Notice there has to be a solution for <math>x</math> between <math>(2,-3)</math> and <math>(1,2)</math> because of the floors. There is also no way <math>2</math> solutions exist because of the quadratic, and when we add them together, we get <math>\boxed{(\text{B}) \ 4}.</math> | |
− | ~ | + | ~perion, |
+ | minor grammar edit by Ynsg. | ||
− | ==Solution | + | ==Solution 3 (Three Cases)== |
First, let's take care of the integer case--clearly, only <math>x=1,2</math> work. | 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>. | 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>. | ||
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~wuwang2002 | ~wuwang2002 | ||
− | == Solution | + | == Solution 4== |
First, <math>x=2,1</math> are trivial solutions | First, <math>x=2,1</math> are trivial solutions | ||
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~kjljixx | ~kjljixx | ||
− | ==Solution | + | ==Solution 5== |
Denote <math>a = \lfloor x \rfloor</math>. | Denote <math>a = \lfloor x \rfloor</math>. | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | == Solution | + | == Solution 6 (Quick) == |
A quadratic equation 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. | A quadratic equation 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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~Technodoggo | ~Technodoggo | ||
− | == Solution | + | == Solution 7 == |
<math>x=1, 2</math> are trivial solutions. | <math>x=1, 2</math> are trivial solutions. | ||
Let <math>x=n+f</math> for some integer <math>n</math> and some number <math>f</math> such that <math>-1<f<1</math>. <cmath>\lfloor{x}\rfloor^2-3x+2= \lfloor{n+f}\rfloor^2-3(n+f)+2=n^2+-3(n+f)+2.</cmath> So now we have | Let <math>x=n+f</math> for some integer <math>n</math> and some number <math>f</math> such that <math>-1<f<1</math>. <cmath>\lfloor{x}\rfloor^2-3x+2= \lfloor{n+f}\rfloor^2-3(n+f)+2=n^2+-3(n+f)+2.</cmath> So now we have | ||
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~azc1027 | ~azc1027 | ||
− | ==Solution | + | ==Solution 8== |
− | |||
− | |||
− | |||
We rewrite the equation as <math>{\lfloor x\rfloor}^2-3\lfloor x\rfloor-3\{x\}+2=0</math>, where <math>\{x\}</math> is the fractional part of <math>x</math> | We rewrite the equation as <math>{\lfloor x\rfloor}^2-3\lfloor x\rfloor-3\{x\}+2=0</math>, where <math>\{x\}</math> is the fractional part of <math>x</math> | ||
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Solving, we have <math> | Solving, we have <math> | ||
− | \left[\frac{3-\sqrt{13}}{2},1\right]\cup \left[2,\frac{3+\sqrt{13}}{2}\right]</math>. But since <math> | + | \left[\frac{3-\sqrt{13}}{2},1\right]\cup \left[2,\frac{3+\sqrt{13}}{2}\right]</math>. But since <math>x_1</math> is an integer, we have <math>x_1</math> can only be <math>0,1,2,</math> or <math>3</math>. |
− | Testing, we see these values of <math> | + | Testing, we see these values of <math>x_1</math> work, and therefore the answer is just <math>\boxed{\textbf{(B) }4}</math>. |
~ESAOPS | ~ESAOPS | ||
+ | ==Similar approach as Solution 8== | ||
+ | Use the fact that <math>x = \lfloor x \rfloor + \{x\}</math>. Thus we have | ||
+ | <cmath>(\lfloor x \rfloor^2 - 3\lfloor x \rfloor + 2) - 3\{x\} = 0.</cmath> | ||
− | + | Noting that <math>0 \leq \{x\} < 1</math>, we get | |
− | + | <cmath>0 \leq (\lfloor x \rfloor - 2)(\lfloor x \rfloor - 1) < 3.</cmath> | |
− | ~ | + | From there, it is not too hard to see that the only values of <math>\lfloor x \rfloor</math> that satisfy this condition (while also noting that <math>\lfloor x \rfloor</math> must be an integer) are 3, 2, 1, and 0, yielding 4 values. |
+ | |||
+ | ~mathboy282 | ||
==Solution 9 (Very Fast)== | ==Solution 9 (Very Fast)== | ||
− | We know that for integer values of x, the graph is just <math>x^2-3x+2</math>. From the interval <math>[x, x+1 | + | We know that for integer values of x, the graph is just <math>x^2-3x+2</math>. From the interval <math>[x, x+1]</math>, 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 <math>x = 0, 1, 2, 3</math> results in a <math>x^2-3x+2</math> in the interval <math>[0, 3]</math>.That is <math>\boxed{\textbf{(B) }4}</math> solutions. |
~Xyco | ~Xyco | ||
− | ==Solution 10 == | + | ==Solution 10== |
− | |||
− | |||
− | |||
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− | + | Define <math>[x] = n,</math> and define the fractional part of <math>x</math> as <math>x(f).</math> | |
− | <math>n^2-3(n+x(f))+2=0</math> | + | Thus <math>\lfloor{x}\rfloor^2-3x+2=0</math> is <cmath>n^2-3(n+x(f))+2=0.</cmath> Expanding, <cmath>n^2-3n-3x(f)+2=0.</cmath> We realize <math>n^2-3n+2</math> must always be an integer so for <math>n^2-3n+2</math> to equal zero, <math>3x(f)</math> must also equal an integer. Therefore, <math>x(f)</math> must be <math>\frac{1}{3}, \frac{2}{3},</math> or <math>0.</math> Plugging in <math>x(f)=0</math> gives <cmath>n^2-3n-3(0)+2=0,</cmath> which simplifies to <cmath>n^2-3n+2=0.</cmath> Continuing, plugging in <math>x(f)=\frac{1}{3}</math> gives <cmath>n^2-3n-3\left(\frac{1}{3}\right)+2=0,</cmath> which simplifies to <cmath>n^2-3n+1.</cmath> Finally, substituting <math>x(f)=\frac{2}{3}</math> gives <cmath>n^2-3n-3\left(\frac{2}{3}\right)+2=0,</cmath> which simplifies to <cmath>n^2-3n=0.</cmath> We know <math>n</math> must be a integer, so we can just solve for <math>n</math> and only utilize the integers we end up with. |
− | <math>n | + | We get two integers <math>n</math> if <math>x(f)=0:</math> <math>n=1,2.</math> We know <math>n+x(f)=x</math> by definition, and plugging the values of <math>n</math> and <math>x(f)</math> into this equation gives us two solutions for <math>x:</math> <cmath>x=1,2.</cmath> |
− | <math>n^2- | + | We don't get any integers <math>n</math> if <math>x(f)=\frac{1}{3}.</math> We use the quadratic discriminant, <math>\sqrt{b^{2}-4ac}</math> to get an expression which yields a non-integer value, <math>\sqrt{3^{2}-4(1)(1)},</math> which means this case is invalid. |
− | |||
− | <math> | ||
− | + | We get two integers <math>n</math> if <math>x(f)=\frac{2}{3}:</math> <math>n=0,3.</math> We know <math>n+x(f)=x</math> by definition, and plugging the values of <math>n</math> and <math>x(f)</math> into this equation gives us two solutions for <math>x:</math> <cmath>x=\frac{2}{3},\frac{11}{3}.</cmath> | |
− | |||
− | + | We end up with a total of four solutions which are <math>x=1,2, \frac{2}{3}, \frac{11}{3}.</math> | |
− | + | Our answer is | |
− | < | + | <cmath>\boxed{\textbf{(B) }4}</cmath> |
− | |||
− | |||
− | + | ~formatting by belindazhu13 | |
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− | |||
− | |||
− | |||
− | + | ==Solution 11 (Based on graph)== | |
− | + | [[File:2023AMC10BQ22Solution.jpg|400px]] | |
− | <math> | + | For <math>\lfloor x \rfloor^2 - 3x + 2 = 0</math>, there is a discontinuity at each integer value of <math>x</math>, and it also lies on the non-floor version of the function. Between each integer <math>x</math> and the next forms a line with a slope of <math>-3</math>. This simplifies the task of sketching the function's graph. Note that the points at <math>1</math> and <math>2</math> are considered intersections because they are points on the left side of each integer interval that exactly lie on <math>y = 0</math>. Thus, we conclude there are <math>4</math> intersection points, and the answer is <math>\boxed{(\text{B}) \ 4}</math>. |
− | <math> | ||
− | <math> | ||
− | <math>2 | ||
− | + | ~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx] | |
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== |
Latest revision as of 13:40, 1 December 2024
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Desperation)
- 4 Solution 3 (Three Cases)
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6 (Quick)
- 8 Solution 7
- 9 Solution 8
- 10 Similar approach as Solution 8
- 11 Solution 9 (Very Fast)
- 12 Solution 10
- 13 Solution 11 (Based on graph)
- 14 Video Solution 1 by OmegaLearn
- 15 Video Solution 2 by SpreadTheMathLove
- 16 Video Solution
- 17 See also
Problem
How many distinct values of satisfy
, where
denotes the largest integer less than or equal to
?
Solution 1
To further grasp at this equation, we rearrange the equation into
Thus,
is a perfect square and nonnegative. It is now much more apparent that
and that
is a solution.
Additionally, by observing the RHS, as
since squares grow quicker than linear functions.
Now that we have narrowed down our search, we can simply test for intervals This intuition to use intervals stems from the fact that
are observable integral solutions.
Notice how there is only one solution per interval, as increases while
stays the same.
Finally, we see that does not work, however, through setting
is a solution and within our domain of
This provides us with solutions thus the final answer is
~mathbrek, happyhari
Solution 2 (Desperation)
Notice there has to be a solution for between
and
because of the floors. There is also no way
solutions exist because of the quadratic, and when we add them together, we get
~perion,
minor grammar edit by Ynsg.
Solution 3 (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
.
~wuwang2002
Solution 4
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 5
Denote .
Denote
.
Thus,
.
The equation given in this problem can be written as
Thus,
Because , we have
.
Thus,
If ,
so
can be
.
If ,
which we find has no integer solutions after finding the discriminant.
If ,
->
so
can also be
.
Therefore, , 2, 0, 3.
Therefore, the number of solutions is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 6 (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 7
are trivial solutions.
Let
for some integer
and some number
such that
.
So now we have
which we can rewrite as
Since
is an integer,
is an integer, so
is an integer. Since
, the only possible values of
are
,
,
, and
. Plugging in each value, we find that the only value of
that produces integer solutions for
is
. If
,
or
. Hence, there is a total of 4 possible solutions, so the answer is
.
~azc1027
Solution 8
We rewrite the equation as , where
is the fractional part of
Denote and
Thus
By definition, . We then have
and therefore
.
Solving, we have . But since
is an integer, we have
can only be
or
.
Testing, we see these values of work, and therefore the answer is just
.
~ESAOPS
Similar approach as Solution 8
Use the fact that . Thus we have
Noting that , we get
From there, it is not too hard to see that the only values of that satisfy this condition (while also noting that
must be an integer) are 3, 2, 1, and 0, yielding 4 values.
~mathboy282
Solution 9 (Very Fast)
We know that for integer values of x, the graph is just . From the interval
, 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
results in a
in the interval
.That is
solutions.
~Xyco
Solution 10
Define and define the fractional part of
as
Thus is
Expanding,
We realize
must always be an integer so for
to equal zero,
must also equal an integer. Therefore,
must be
or
Plugging in
gives
which simplifies to
Continuing, plugging in
gives
which simplifies to
Finally, substituting
gives
which simplifies to
We know
must be a integer, so we can just solve for
and only utilize the integers we end up with.
We get two integers if
We know
by definition, and plugging the values of
and
into this equation gives us two solutions for
We don't get any integers if
We use the quadratic discriminant,
to get an expression which yields a non-integer value,
which means this case is invalid.
We get two integers if
We know
by definition, and plugging the values of
and
into this equation gives us two solutions for
We end up with a total of four solutions which are
Our answer is
~formatting by belindazhu13
Solution 11 (Based on graph)
For , there is a discontinuity at each integer value of
, and it also lies on the non-floor version of the function. Between each integer
and the next forms a line with a slope of
. This simplifies the task of sketching the function's graph. Note that the points at
and
are considered intersections because they are points on the left side of each integer interval that exactly lie on
. Thus, we conclude there are
intersection points, and the answer is
.
Video Solution 1 by OmegaLearn
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=DvHGEXBjf0Y
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.