Difference between revisions of "2020 AIME I Problems/Problem 11"
(→Solution) |
|||
Line 19: | Line 19: | ||
-EZmath2006 | -EZmath2006 | ||
+ | |||
+ | == 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 12:48, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.