Difference between revisions of "2018 AMC 10B Problems/Problem 25"
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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>. (Mudkipswims42) | 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>. (Mudkipswims42) | ||
+ | |||
+ | == Alternative Solution (Please excuse me for bad latex) == | ||
+ | Same as the first solution, x^2=10,000{x}. | ||
+ | |||
+ | We can write x as floor(x)+{x}. Expanding everything, we get a quadratic in {x} in terms of floor(x): | ||
+ | |||
+ | {x}^2+ (2*floor(x)-10,000){x} + [floor(x)]^2 = 0 | ||
+ | |||
+ | We use the quadratic formula to solve for {x}. | ||
+ | {x} = [ -2*floor(x)+10,000 +/- sqrt( [-2*floor(x)+10,000]^2- 4*[floor(x)]^2 ) ]/2 | ||
+ | |||
+ | Since 0=<{x}<1, we get an inequality which we can then solve. After simplifying a lot, we get that [floor(x)]^2 + 2*floor(x) - 9999 < 0 | ||
+ | |||
+ | Solving over the integers, -101< floor(x) < 99, and since floor(x) is an integer, there are 199 solutions. Each value of floor(x) should correspond to one value of x, so we are done. | ||
+ | |||
+ | ~Alex_z_Awesome | ||
+ | This solution needs some editing. Please edit if you have time. | ||
==See Also== | ==See Also== |
Revision as of 18:27, 17 February 2018
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
. (Mudkipswims42)
Alternative Solution (Please excuse me for bad latex)
Same as the first solution, x^2=10,000{x}.
We can write x as floor(x)+{x}. Expanding everything, we get a quadratic in {x} in terms of floor(x):
{x}^2+ (2*floor(x)-10,000){x} + [floor(x)]^2 = 0
We use the quadratic formula to solve for {x}. {x} = [ -2*floor(x)+10,000 +/- sqrt( [-2*floor(x)+10,000]^2- 4*[floor(x)]^2 ) ]/2
Since 0=<{x}<1, we get an inequality which we can then solve. After simplifying a lot, we get that [floor(x)]^2 + 2*floor(x) - 9999 < 0
Solving over the integers, -101< floor(x) < 99, and since floor(x) is an integer, there are 199 solutions. Each value of floor(x) should correspond to one value of x, so we are done.
~Alex_z_Awesome This solution needs some editing. Please edit if you have time.
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.