Difference between revisions of "2020 AIME I Problems/Problem 11"
(→Solution 1 (Strategic Casework)) |
m (→Notes For *: For bold, italic, underlining, or Asymptote, instead of using brackets, like usual, use <> instead, but only works here.) |
||
Line 21: | Line 21: | ||
== Notes For * == | == Notes For * == | ||
− | In case anyone is confused by this (as I initially was). In the case where <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. So basically since <math>a=-6</math> in this case, <math>f(2)=f(4)=b-8</math>, and we have <math>21</math> choices for b and we | + | In case anyone is confused by this (as I initially was). In the case where <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. So basically since <math>a=-6</math> in this case, <math>f(2)=f(4)=b-8</math>, and we have <math>21</math> choices for b and we <i>still can</i> 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 <math>b-8</math> ensures this, and of course an integer multiplied by an integer is an integer so <math>d</math> will still be an integer. In other words, you have can have <math>b</math> and <math>c</math> be any integer with absolute value less than or equal to 10 with <math>d</math> still being an integer. Now refer back to the 1st solution. |
~First | ~First | ||
Revision as of 14:43, 22 May 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 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
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.