Difference between revisions of "2020 AMC 12A Problems/Problem 25"
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Let <math>f(x)=\lfloor x \rfloor \cdot \{x\}</math> and <math>g(x)=a \cdot x^2.</math> | Let <math>f(x)=\lfloor x \rfloor \cdot \{x\}</math> and <math>g(x)=a \cdot x^2.</math> | ||
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Graph in Desmos: https://www.desmos.com/calculator/ouvaiqjdzj | Graph in Desmos: https://www.desmos.com/calculator/ouvaiqjdzj | ||
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~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 09:35, 19 May 2021
Contents
[hide]Problem
The number , where
and
are relatively prime positive integers, has the property that the sum of all real numbers
satisfying
is
, where
denotes the greatest integer less than or equal to
and
denotes the fractional part of
. What is
?
Solution 1
Let be the unique solution in this range. Note that
is also a solution as long as
, hence all our solutions are
for some
. This sum
must be between
and
, which gives
and
. Plugging this back in gives
.
Solution 2
First note that when
while
. Thus we only need to look at positive solutions (
doesn't affect the sum of the solutions).
Next, we breakdown
down for each interval
, where
is a positive integer. Assume
, then
. This means that when
,
. Setting this equal to
gives
We're looking at the solution with the positive
, which is
. Note that if
is the greatest
such that
has a solution, the sum of all these solutions is slightly over
, which is
when
, just under
. Checking this gives
~ktong
Remarks
Let and
We make the following table of values:
We graph (in red, by branches) and
(in blue, for
) as shown below.
Graph in Desmos: https://www.desmos.com/calculator/ouvaiqjdzj
~MRENTHUSIASM
Video Solution 1 (Geometry)
This video shows how things like The Pythagorean Theorem and The Law of Sines work together to solve this seemingly algebraic problem: https://www.youtube.com/watch?v=6IJ7Jxa98zw&feature=youtu.be
Video Solution 2
https://www.youtube.com/watch?v=xex8TBSzKNE ~ MathEx
Video Solution 3 (by Art of Problem Solving)
https://www.youtube.com/watch?v=7_mdreGBPvg&t=428s&ab_channel=ArtofProblemSolving
Created by Richard Rusczyk
See Also
2020 AMC 12A (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.