Difference between revisions of "2018 AMC 10B Problems/Problem 25"
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There are 199 elements in this range, so the answer is <math>\fbox{C 199}</math> | There are 199 elements in this range, so the answer is <math>\fbox{C 199}</math> | ||
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− | As in | + | As in Solution 1, we can write <math>x^2=10,000\{x\}</math>. Obviously, <math>0<\{x\}<1</math>, so x is between <math>(-100,100)</math>. Hence, the answer is <math>\boxed{199}</math>. |
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− | Hence, the answer is <math>\boxed{199}</math>. | ||
-Pi_3.14_Squared | -Pi_3.14_Squared |
Revision as of 19:45, 23 February 2018
Contents
[hide]Problem
Let denote the greatest integer less than or equal to
. How many real numbers
satisfy the equation
?
Solution
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
.
Alternative, Bashy Solution
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 {x}:
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.
Another Solution
Let where
is the integer portion of
and
is the decimal portion.
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 199 elements in this range, so the answer is
Solution 4
As in Solution 1, we can write . Obviously,
, so x is between
. Hence, the answer is
.
-Pi_3.14_Squared
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.