2023 AMC 10B Problems/Problem 22
Contents
- 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(has 3 cases)
- 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 because of the quadratic, and when we add them together, we get ~perion.
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(has 3 cases)
define
define the fractional part of as
thus is
must always be an 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 simplify into
where n must be a integer
just use solve for n and use only integers
we get 2 integers for the first for
we get 2 integers for the second for
We get ZERO integers for the third for
we get use the quadratic discriminant to see
Our equation is
yielding a non integer value which means this case is invalid
we count a total of 4 solutions which are
Our answer is
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 . ～Athmyx
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 |
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