Difference between revisions of "2020 AIME I Problems/Problem 11"
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− | + | There can be two different cases for this problem, either <math>f(2)=f(4)</math> or not. If it is, note that Vieta's forces <math>a = -6</math>. Then, <math>b</math> can be anything. However, <math>c</math> can also be anything, as we can set the root of <math>g</math> (not equal to <math>f(2) = f(4)</math>) to any integer, producing a possible integer value of <math>d</math>. Therefore there are <math>21^2 = 441</math> in this case*. If it isn't, then <math>f(2),f(4)</math> are the roots of <math>g</math>. This means by Vieta's, that: | |
<cmath>f(2)+f(4) = -c \in [-10,10]</cmath> | <cmath>f(2)+f(4) = -c \in [-10,10]</cmath> |
Revision as of 13:42, 18 January 2024
Contents
[hide]Problem
For integers and
let
and
Find the number of ordered triples
of integers with absolute values not exceeding
for which there is an integer
such that
Solution 1 (Strategic Casework)
There can be two different cases for this problem, either or not. If it is, note that Vieta's forces
. Then,
can be anything. However,
can also be anything, as we can set the root of
(not equal to
) to any integer, producing a possible integer value of
. Therefore there are
in this case*. If it isn't, then
are the roots of
. This means by Vieta's, that:
Solving these inequalities while considering that to prevent
, we obtain
possible tuples and adding gives
.
~awang11
Solution 2 (Bash)
Define . Since
, we know
. Plugging in
into
, we get
. Setting
,
. Simplifying and cancelling terms,
Therefore, either or
. The first case is easy:
and there are
tuples in that case. In the second case, we simply perform casework on even values of
, to get
tuples, subtracting the
tuples in both cases we get
.
-EZmath2006
Solution 3 (Official MAA)
For a given ordered triple , the value of
is uniquely determined, and a value of
can be found to give
any prescribed integer value. Hence the required condition can be satisfied provided that
,
, and
are chosen so that
First suppose that
, so
. In this case there are
choices for each of
and
with
and
, so this case accounts for
ordered triples.
Next suppose that and
, so
. Because
, it follows that
, and because
, it follows that
. Then
. The number of ordered triples for various values of
are presented in the following table.
The total number of ordered triples that satisfy the required condition is
.
Notes For *
In case anyone is confused by this (as I initially was). In the case where , this does not mean that g has a double root of
, ONLY that
is one of the roots of g. So basically since
in this case,
, and we have
choices for b and we still can ensure c is an integer with absolute value less than or equal to 10 simply by having another integer root of g that when added to
ensures this, and of course an integer multiplied by an integer is an integer so
will still be an integer. In other words, you have can have
and
be any integer with absolute value less than or equal to 10 with
still being an integer. Now refer back to the 1st solution.
~First
Video Solution
https://www.youtube.com/watch?v=ftqYFzzWKv8&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=8 ~ MathEx
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.