Difference between revisions of "2020 AIME I Problems/Problem 11"
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== Problem == | == Problem == | ||
Revision as of 16:47, 12 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
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:
![\[f(2)+f(4) = -c \in [-10,10]\]](http://latex.artofproblemsolving.com/6/a/3/6a37efd5cf28128ffc86292858c5295324219916.png)
![\[20 + 6a + 2b \in [-10,10]\]](http://latex.artofproblemsolving.com/6/a/2/6a21e367c68e23d262ef4c634a2cac27eadf144c.png)
![\[3a + b \in [-15,5].\]](http://latex.artofproblemsolving.com/9/5/5/95555d558a85629834280abcce4f6c78c0a6cc4c.png)
Solving these inequalities while considering that 
to prevent , we obtain 
possible tuples and adding gives 
. ~awang11
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. 
