Difference between revisions of "2024 AMC 12B Problems/Problem 17"

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~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
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==See also==
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{{AMC12 box|year=2024|ab=B|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 00:55, 14 November 2024

Problem 17

Integers $a$ and $b$ are randomly chosen without replacement from the set of integers with absolute value not exceeding $10$. What is the probability that the polynomial $x^3 + ax^2 + bx + 6$ has $3$ distinct integer roots?

$\textbf{(A)} \frac{1}{240} \qquad \textbf{(B)} \frac{1}{221} \qquad \textbf{(C)} \frac{1}{105} \qquad \textbf{(D)} \frac{1}{84} \qquad \textbf{(E)} \frac{1}{63}$.

Solution

Solution 1

-10<= a, b <= 10 , a,b has 21 choices per Vieta, x1x2x3 = -6, x1 + x2+ x3 = -a , x1x2+ x2x3 + x3x1 = b

Case: (1) (x1,x2,x3) = (-1,-1,6) , b = 13 not valid

(2) (x1,x2,x3) = (-1,1,6) , b = -1, a=-6 valid

(3) (x1,x2,x3) = ( 1,2,-3) , b = -7, a=0 valid

(4) (x1,x2,x3) = (1,-2,3) , b = -7, a=2 valid

(5) (x1,x2,x3) = (-1,2,3) , b = 1, a=4 valid

(6) (x1,x2,x3) = (-1,-2,-3) , b = 11 invalid

probability = $\frac{4}{21*20}$ = $\frac{1}{105}$ $\boxed{\textbf{(C) }\frac{1}{105}}$

~luckuso

See also

2024 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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