2023 AMC 10B Problems/Problem 22
Contents
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 the 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 (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
We rewrite the equation as 3{}, where {} 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.