2018 AMC 10B Problems/Problem 25
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.