Difference between revisions of "2018 AMC 10B Problems/Problem 25"
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+ | {{duplicate|[[2018 AMC 10B Problems#Problem 25 | 2018 AMC 10B #25]] and [[2018 AMC 12B Problems#Problem 24|2018 AMC 12B #24]]}} | ||
+ | |||
== Problem == | == Problem == | ||
Let <math>\lfloor x \rfloor</math> denote the greatest integer less than or equal to <math>x</math>. How many real numbers <math>x</math> satisfy the equation <math>x^2 + 10,000\lfloor x \rfloor = 10,000x</math>? | Let <math>\lfloor x \rfloor</math> denote the greatest integer less than or equal to <math>x</math>. How many real numbers <math>x</math> satisfy the equation <math>x^2 + 10,000\lfloor x \rfloor = 10,000x</math>? | ||
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<math>\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201</math> | <math>\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201</math> | ||
− | == Solution == | + | ==Solution 1== |
This rewrites itself to <math>x^2=10,000\{x\}</math>. | This rewrites itself to <math>x^2=10,000\{x\}</math>. | ||
Graphing <math>y=10,000\{x\}</math> and <math>y=x^2</math> we see that the former is a set of line segments with slope <math>10,000</math> from <math>0</math> to <math>1</math> with a hole at <math>x=1</math>, then <math>1</math> to <math>2</math> with a hole at <math>x=2</math> etc. | Graphing <math>y=10,000\{x\}</math> and <math>y=x^2</math> we see that the former is a set of line segments with slope <math>10,000</math> from <math>0</math> to <math>1</math> with a hole at <math>x=1</math>, then <math>1</math> to <math>2</math> with a hole at <math>x=2</math> etc. | ||
− | |||
Here is a graph of <math>y=x^2</math> and <math>y=16\{x\}</math> for visualization. | Here is a graph of <math>y=x^2</math> and <math>y=16\{x\}</math> for visualization. | ||
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Now notice that when <math>x=\pm 100</math> then graph has a hole at <math>(\pm 100,10,000)</math> which the equation <math>y=x^2</math> passes through and then continues upwards. Thus our set of possible solutions is bounded by <math>(-100,100)</math>. We can see that <math>y=x^2</math> intersects each of the lines once and there are <math>99-(-99)+1=199</math> lines for an answer of <math>\boxed{\text{(C)}~199}</math>. | Now notice that when <math>x=\pm 100</math> then graph has a hole at <math>(\pm 100,10,000)</math> which the equation <math>y=x^2</math> passes through and then continues upwards. Thus our set of possible solutions is bounded by <math>(-100,100)</math>. We can see that <math>y=x^2</math> intersects each of the lines once and there are <math>99-(-99)+1=199</math> lines for an answer of <math>\boxed{\text{(C)}~199}</math>. | ||
− | == | + | ==Solution 2== |
Same as the first solution, <math>x^2=10,000\{x\} </math>. | Same as the first solution, <math>x^2=10,000\{x\} </math>. | ||
− | We can write <math>x</math> as <math>\lfloor x \rfloor+\{x\}</math>. Expanding everything, we get a quadratic in <math>{x}</math> in terms of <math>\lfloor x \rfloor</math>: | + | We can write <math>x</math> as <math>\lfloor x \rfloor+\{x\}</math>. Expanding everything, we get a quadratic in <math>\{x\}</math> in terms of <math>\lfloor x \rfloor</math>: |
− | < | + | <cmath> \{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0</cmath> |
− | We use the quadratic formula to solve for {x}: | + | We use the quadratic formula to solve for <math>\{x\}</math> : |
− | < | + | <cmath> \{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ ( 2\lfloor x \rfloor - 10,000 )^2 - 4\lfloor x \rfloor^2 }}{2} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ 4\lfloor x \rfloor^2 -40,000 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 }}{2} </cmath> |
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− | Solving over the integers, <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of<math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done. | + | Solving over the integers, <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of <math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done. |
− | |||
− | |||
− | == | + | ==Solution 3== |
− | Let <math>x = a+k</math> where <math>a</math> is the integer | + | Let <math>x = a+k</math> where <math>a</math> is the integer part of <math>x</math> and <math>k</math> is the fractional part of <math>x</math>. |
We can then rewrite the problem below: | We can then rewrite the problem below: | ||
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<math>(a+k)^2 + 10000a = 10000a + 10000k</math> | <math>(a+k)^2 + 10000a = 10000a + 10000k</math> | ||
− | Solving for a+k | + | Solving for <math>a+k = x</math> |
<math>(a+k)^2 = 10000k</math> | <math>(a+k)^2 = 10000k</math> | ||
− | <math>a+k = \pm100\sqrt{k}</math> | + | <math>x = a+k = \pm100\sqrt{k}</math> |
+ | |||
+ | Because <math>0 \leq k < 1</math>, we know that <math>a+k</math> cannot be less than or equal to <math>-100</math> nor greater than or equal to <math>100</math>. Therefore: | ||
+ | |||
+ | <math>-99 \leq x \leq 99</math> | ||
+ | |||
+ | There are <math>199</math> elements in this range, so the answer is <math>\boxed{\textbf{(C)} \text{ 199}}</math>. | ||
+ | |||
+ | Note (not by author): this solution seems to be invalid at first, because one can not determine whether <math>x</math> is an integer or not. However, it actually works because although <math>x</math> itself might not be an integer, it is very close to one, so there are 199 potential <math>x</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Notice the given equation is equivilent to <math>(\lfloor x \rfloor+\{x\})^2=10,000\{x\} </math> | ||
+ | |||
+ | Now we now that <math>\{x\} < 1</math> so plugging in <math>1</math> for <math>\{x\}</math> we can find the upper and lower bounds for the values. | ||
+ | |||
+ | <math>(\lfloor x \rfloor +1)^2 = 10,000(1)</math> | ||
+ | |||
+ | <math>(\lfloor x \rfloor +1) = \pm 100</math> | ||
+ | |||
+ | <math>\lfloor x \rfloor = 99, -101</math> | ||
+ | |||
+ | And just like <math>\textbf{Solution 2}</math>, we see that <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of <math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done. | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | Firstly, if <math>x</math> is an integer, then <math>10,000\lfloor x \rfloor=10,000x</math>, so <math>x</math> must be <math>0</math>. | ||
+ | |||
+ | If <math>0<x<1</math>, then we know the following: | ||
+ | |||
+ | <math>0<x^2<1</math> | ||
+ | |||
+ | <math>10,000\lfloor x \rfloor =0</math> | ||
+ | |||
+ | <math>0<10,000x<10,000</math> | ||
+ | |||
+ | Therefore, <math>0<x^2+10,000\lfloor x \rfloor <1</math>, which overlaps with <math>0<10,000x<10,000</math>. This means that there is at least one real solution between <math>0</math> and <math>1</math>. Since <math>x^2+10,000\lfloor x \rfloor </math> increases quadratically and <math>10,000x</math> increases linearly, there is only one solution for this case. | ||
+ | |||
+ | Similarly, if <math>1<x<2</math>, then we know the following: | ||
+ | |||
+ | <math>1<x^2<4</math> | ||
+ | |||
+ | <math>10,000\lfloor x \rfloor =10,000</math> | ||
+ | |||
+ | <math><10,000<10,000x<20,000</math> | ||
+ | |||
+ | By following similar logic, we can find that there is one solution between <math>1</math> ad <math>2</math>. | ||
+ | |||
+ | We can also follow the same process to find that there are negative solutions for <math>x</math> as well. | ||
+ | |||
+ | There are not an infinite amount of solutions, so at one point there will be no solutions when <math>n<x<n+1</math> for some integer <math>n</math>. For there to be no solutions in a given range means that the range of <math>10,000\lfloor x \rfloor + x^2</math> does not intersect the range of <math>10,000x</math>. <math>x^2</math> will always be positive, and <math>10,000\lfloor x \rfloor</math> is less than <math>10,000</math> less than <math>10,000x</math>, so when <math>x^2 >= 10,000</math>, the equation will have no solutions. This means that there are <math>99</math> positive solutions, <math>99</math> negative solutions, and <math>0</math> for a total of <math>\boxed{\text{(C)}~199}</math> solutions. | ||
+ | |||
+ | ~Owen1204 | ||
+ | |||
+ | ==Solution 6 (General Equation)== | ||
+ | |||
+ | General solution to this type of equation <math>f(x, \lfloor x \rfloor) = 0</math>: | ||
+ | |||
+ | 1. solve <math>f(x, \lfloor x \rfloor) = 0</math> for <math>x</math> to get <math>x = g(\lfloor x \rfloor )</math> | ||
+ | 2. apply <math>\lfloor x \rfloor \le x < \lfloor x \rfloor+1</math>, solve <math>\lfloor x \rfloor \le g(\lfloor x \rfloor) < \lfloor x \rfloor+1</math> to get the domain of <math>\lfloor x \rfloor</math> | ||
+ | 3. get <math> \lfloor x \rfloor</math> from the domain of <math> \lfloor x \rfloor</math> because <math> \lfloor x \rfloor</math> is integer, then get <math>x</math> from <math> \lfloor x \rfloor</math> by <math>x = g( \lfloor x \rfloor) </math> | ||
+ | Note: function <math>\lfloor x \rfloor</math> maps <math>x</math> to its floor. By solving <math>f(x, \lfloor x \rfloor) = 0</math>, we get function <math>x = g( \lfloor x \rfloor) </math>, mapping <math>x</math>'s floor to <math>x</math> | ||
+ | |||
+ | <math>x^2 - 10000x + 10000 \lfloor x \rfloor =0</math> | ||
− | + | <math>x=5000 \pm 100 \sqrt{2500- \lfloor x \rfloor}</math>, <math>\lfloor x \rfloor \le 2500</math> | |
− | <math> | + | <math>\lfloor x \rfloor \le x < \lfloor x \rfloor + 1</math> |
− | + | If <math>x= 5000 + 100 \sqrt{2500 - \lfloor x \rfloor}</math>, <math>x \ge 5000</math>, it contradicts <math>x < \lfloor x \rfloor + 1 \le 2501</math> | |
+ | So <math>x= 5000 - 100 \sqrt{2500 - \lfloor x \rfloor}</math> | ||
− | + | Let <math>k = \lfloor x \rfloor</math> , <math>x= 5000 - 100 \sqrt{2500 - k}</math> | |
+ | |||
+ | <math>k \le 5000 - 100 \sqrt{2500 - k} < k + 1</math> | ||
+ | |||
+ | <math>0 \le 5000 - k - 100 \sqrt{2500 - k} < 1</math> | ||
+ | |||
+ | <math>0 \le 2500 - k - 100 \sqrt{2500 - k} + 2500 < 1</math> | ||
+ | |||
+ | <math>0 \le (\sqrt{2500 - k} - 50)^2 < 1</math> | ||
+ | |||
+ | <math>-1 < \sqrt{2500 - k} - 50 < 1</math> | ||
+ | |||
+ | <math>49 < \sqrt{2500 - k} < 51</math> | ||
+ | |||
+ | <math>-101 < k < 99</math> | ||
+ | |||
+ | So the number of <math>k</math>'s values is <math>99-(-101)-1=199</math>. Because <math>x=5000-100\sqrt{2500-k}</math>, for each value of <math>k</math>, there is a value for <math>x</math>. The answer is <math>\boxed{\textbf{(C)} 199}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 7== | ||
+ | Subtracting <math>10000\lfloor x\rfloor</math> from both sides gives <math>x^2=10000(x-\lfloor x\rfloor)=10000\{x\}</math>. Dividing both sides by <math>10000</math> gives <math>\left(\frac{x}{100}\right)^2=\{x\}<1</math>. <math>\left(\frac{x}{100}\right)^2<1</math> when <math>-100<x<100</math> so the answer is <math>\boxed{199}</math> | ||
+ | |||
+ | ~randomdude10807 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=vHKPbaXwJUE | ||
==See Also== | ==See Also== | ||
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{{AMC12 box|year=2018|ab=B|num-b=23|num-a=25}} | {{AMC12 box|year=2018|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 00:44, 6 February 2022
- The following problem is from both the 2018 AMC 10B #25 and 2018 AMC 12B #24, so both problems redirect to this page.
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 .
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 .
Note (not by author): this solution seems to be invalid at first, because one can not determine whether is an integer or not. However, it actually works because although itself might not be an integer, it is very close to one, so there are 199 potential .
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
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 quadratically 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 integer . For there to be no solutions in a given range means that the range of does not intersect the range of . will always be positive, and is less than less than , so when , the equation will have no solutions. This means that there are positive solutions, negative solutions, and for a total of solutions.
~Owen1204
Solution 6 (General Equation)
General solution to this type of equation :
1. solve for to get 2. apply , solve to get the domain of 3. get from the domain of because is integer, then get from by Note: function maps to its floor. By solving , we get function , mapping 's floor to
,
If , , it contradicts
So
Let ,
So the number of 's values is . Because , for each value of , there is a value for . The answer is
Solution 7
Subtracting from both sides gives . Dividing both sides by gives . when so the answer is
~randomdude10807
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.