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 {x} 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>? | ||
Revision as of 22:34, 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(has 3 cases)
- 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(has 3 cases)
define
define the fractional part of as
thus is
must alway be a integer and thus for this equal zero must also equal integer
thus must be fraction
and q must be 0,1,2
thus must be 1/3 or 2/3 or 0/3
plugging in all
we simpfly into
where n must be a integer
just use solve for n and use only integers
we get 2 integrs for the first for
we get 2 integers for teh second for
We get ZERO integers for the third for
we get use the quadratic discriminant to see
Our equation is
yeilding a non integer value which means this case is invalid
we count a total of 4 solutiosn which are
Our 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.