Difference between revisions of "2020 AIME I Problems/Problem 11"
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-EZmath2006 | -EZmath2006 | ||
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+ | == Notes == | ||
+ | In case anyone is confused by this (as I initially was). Say <math>f(2)=f(4)</math>. This does not mean that g has a double root of <math>f(2)=f(4)=c</math>, ONLY that <math>c</math> is one of the roots of g. | ||
+ | ~First | ||
==See Also== | ==See Also== |
Revision as of 11:48, 28 March 2020
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)
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
Notes
In case anyone is confused by this (as I initially was). Say . This does not mean that g has a double root of
, ONLY that
is one of the roots of g.
~First
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 |
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