Difference between revisions of "2023 AMC 10B Problems/Problem 22"
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+ | ==Video Solution== | ||
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+ | https://youtu.be/ONRoop23LIY | ||
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+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See also== | ==See also== | ||
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{{MAA Notice}} | {{MAA Notice}} |
Revision as of 03:02, 20 November 2023
Contents
Problem
How many distinct values of 𝑥 satisfy
, where
denotes the largest integer less than or equal to
?
Solution 1(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 2
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 3
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 4(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 5
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 6
We rewrite the equation as 3{x}
, where {x} is the fractional part of
To rewrite it just for simplicity we rewrite as .
We must have by our 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
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.