2018 AMC 10B Problems/Problem 25
Contents
Problem
Let denote the greatest integer less than or equal to . How many real numbers satisfy the equation ?
Solution 1
This rewrites itself to .
Graphing and we see that the former is a set of line segments with slope from to with a hole at , then to with a hole at etc.
Here is a graph of and for visualization.
Now notice that when then graph has a hole at which the equation passes through and then continues upwards. Thus our set of possible solutions is bounded by . We can see that intersects each of the lines once and there are lines for an answer of .
Note: From the graph, we can clearly see there are solutions on the negative side of the -axis and only on the positive side of the -axis. So the solution really should be from to , which still counts to . A couple of the alternative solutions also seem to have the same flaw.
Solution 2
Same as the first solution, .
We can write as . Expanding everything, we get a quadratic in in terms of :
We use the quadratic formula to solve for :
Since , we get an inequality which we can then solve. After simplifying a lot, we get that .
Solving over the integers, , and since is an integer, there are solutions. Each value of should correspond to one value of , so we are done.
Note (not by author): this solution is invalid, because one can not determine whether is an integer or not.
Solution 3
Let where is the integer part of and is the fractional part of . We can then rewrite the problem below:
From here, we get
Solving for
Because , we know that cannot be less than or equal to nor greater than or equal to . Therefore:
There are elements in this range, so the answer is .
Solution 4
Notice the given equation is equivilent to
Now we now that so plugging in for we can find the upper and lower bounds for the values.
And just like , we see that , and since is an integer, there are solutions. Each value of should correspond to one value of , so we are done.
Solution 5
First, we can let . We know that by definition. We can rearrange the equation to obtain
.
By taking square root on both sides, we obtain (because ). We know since is the fractional part of , it must be that . Thus, may take any value in the interval . Hence, we know that there are potential values for in that range and we are done.
~awesome1st
Solution 6
Firstly, we can rearrange to get
Rearranging, we get
Noticing that , we know that x can only be within the boundaries of and hence, we know that there are potential values.
Solution 7
Firstly, if is an integer, then , so must be .
If , then we know the following:
Therefore, , which overlaps with . This means that there is at least one real solution between and . Since increases exponentially and increases linearly, there is only one solution for this case.
Similarly, if , then we know the following:
By following similar logic, we can find that there is one solution between ad .
We can also follow the same process to find that there are negative solutions for as well.
There are not an infinite amount of solutions, so at one point there will be no solutions when for some positive integer . Looking at the answer solutions, they are near , so it seems logical that the upper and lower limits are around and . Testing this out, we can see that there is a solution for , but not for . Similarly, there is a solution for , but not for . This means that there are positive solutions, negative solutions, and for a total of solutions.
~Owen1204
Video Solution
https://www.youtube.com/watch?v=vHKPbaXwJUE
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 12 Problems and Solutions |
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