Difference between revisions of "2023 AMC 10B Problems/Problem 22"
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define <math>[x] = n</math> | define <math>[x] = n</math> | ||
− | define the fractional part of x as <math> | + | define the fractional part of <math>x</math> as <math>x(f)</math> |
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== |
Revision as of 00:47, 27 November 2023
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (three cases)
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5 (Quick)
- 7 Solution 6
- 8 Solution 7 (Fakesolve)
- 9 Solution 8 (Graph, similar to Solution 1)
- 10 Solution 9 (Very Fast)
- 11 Solution 10
- 12 Video Solution 1 by OmegaLearn
- 13 Video Solution 2 by SpreadTheMathLove
- 14 Video Solution
- 15 See also
Problem
How many distinct values of 𝑥 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
Sol. 1 (Desperate)
Noticing there has to be a sol. for x is between (2-3) and (1,2) (becuase of the floors), there are 2 sol. becuase of the quadratic and 2 sol. becuase of the floors(as said as before). Using this method of thinking, we get 2+2=4 Giving us the sol. of
~Michaellin16
Solution 2 (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 3
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 4
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 5 (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 6
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 7 (Fakesolve)
Note: This sol is technically bogus as the problem does not state that is necessarily an integer. However, these steps ESAOPS take are great for narrowing down
values and understanding the equation. Additionally, it is worth to note that
and
are not valid values of
~mathbrek
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
Solution 8 (Graph, similar to Solution 1)
Note that we can add to both sides to get
Graphing, we get 4 solutions.
~aleyang
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
.
~Xyco
Solution 10
define
define the fractional part of as
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.