Difference between revisions of "2018 AMC 10B Problems/Problem 25"
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Noticing that <math>10,000 = 100^2</math>, we know that x can only be within the boundaries of <math>-100<x<100</math> and hence, we know that there are <math>\boxed{\text{(C)}~199}</math> potential values. | Noticing that <math>10,000 = 100^2</math>, we know that x can only be within the boundaries of <math>-100<x<100</math> and hence, we know that there are <math>\boxed{\text{(C)}~199}</math> potential values. | ||
+ | |||
+ | ==Solution 7== | ||
+ | |||
+ | 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 exponentially 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 =1</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 positive integer <math>n</math>. Looking at the answer solutions, they are near <math>200</math>, so it seems logical that the upper and lower limits are around <math>100</math> and <math>-100</math>. Testing this out, we can see that there is a solution for <math>99<x<100</math>, but not for <math>100<x<101</math>. Similarly, there is a solution for <math>-100<x<-99</math>, but not for <math>-101<x<-100</math>. 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. | ||
==Video Solution== | ==Video Solution== |
Revision as of 19:05, 28 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
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.
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.