Difference between revisions of "2023 AMC 12B Problems/Problem 6"
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| − | + | ==Solution 1== | |
| + | Looking at the polynomial, we can use Vieta's formulas. Let's say the roots are r, s, and t. Then | ||
| + | <cmath>r+s+t = -a</cmath> | ||
| + | <cmath>rs+rt+st = b</cmath> | ||
| + | <cmath>rst = 6</cmath> | ||
| + | We know that r, s, and t are integers, so we can list the possibilities for distinct solutions using the third formula. We find that <math>(r,s,t)</math> can be <math>(1,2,3), (-1,-2,3), (-1,2,-3), (1,2,-3),</math> and <math>(1,-1,-6)</math>. Remember that r, s, and t have to be different. That's 5 solutions. We can also use those values to find the values of a and b to check that none of them are distinct, and we'll find that they are <math>(6,11), (0,-7), (-2,-5), (-4,1),</math> and <math>(-6,-1)</math>. This means that there are <math>\boxed{5}</math> | ||
Revision as of 18:49, 15 November 2023
Solution 1
Looking at the polynomial, we can use Vieta's formulas. Let's say the roots are r, s, and t. Then
![\[r+s+t = -a\]](http://latex.artofproblemsolving.com/5/2/8/5285652525ea5dd27eb9050321e98bfc1d0eec48.png)
![\[rs+rt+st = b\]](http://latex.artofproblemsolving.com/f/8/4/f84cd6b1f09d7fb99f9c0ad730394cef05610b6f.png)
![\[rst = 6\]](http://latex.artofproblemsolving.com/4/6/f/46fa10a3403c50f0bfcd05efa7802e64b8ad9d4c.png)
We know that r, s, and t are integers, so we can list the possibilities for distinct solutions using the third formula. We find that 
can be 
and 
. Remember that r, s, and t have to be different. That's 5 solutions. We can also use those values to find the values of a and b to check that none of them are distinct, and we'll find that they are 
and 
. This means that there are 
