Difference between revisions of "2018 AMC 10B Problems/Problem 25"
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+ | ==Solution 6== | ||
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+ | Firstly, we can rearrange to get <math>\lfloor x \rfloor = x-x^2/10,000</math> | ||
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+ | Rearranging, we get x^2/10,000 < 1 | ||
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+ | Noticing that 10,000 = 100^2, we know that x can only be within the boundaries of -100<x<100 and hence, we know that there are <math>\boxed{\text{(C)}~199}</math> potential values. | ||
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==Video Solution== | ==Video Solution== | ||
https://www.youtube.com/watch?v=vHKPbaXwJUE | https://www.youtube.com/watch?v=vHKPbaXwJUE |
Revision as of 05:09, 2 January 2020
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.
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 x^2/10,000 < 1
Noticing that 10,000 = 100^2, we know that x can only be within the boundaries of -100<x<100 and hence, we know that there are potential values.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.