Difference between revisions of "2023 AMC 12B Problems/Problem 6"

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<cmath>rst = 6</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>
 
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>
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~lprado

Revision as of 17:49, 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 \[r+s+t = -a\] \[rs+rt+st = b\] \[rst = 6\] 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 $(r,s,t)$ can be $(1,2,3), (-1,-2,3), (-1,2,-3), (1,2,-3),$ and $(1,-1,-6)$. 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 $(6,11), (0,-7), (-2,-5),  (-4,1),$ and $(-6,-1)$. This means that there are $\boxed{5}$

~lprado