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

Revision as of 18:28, 16 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

$-10 \leq a, b \leq 10$ , each of $a,b$ has $21$ choices

Applying Vieta's formulas,

$x_1 \cdot x_2 \cdot x_3 = -6$

$x_1 + x_2+ x_2 = -a$

$x_1 \cdot x_2 + x_1 \cdot x_3 + x_3 \cdot x_2 = b$

Cases:

(1) $(x_1,x_2,x_3) = (-1,1,6) , b = -1, a=-6$ valid

(2) $(x_1,x_2,x_3) = ( 1,2,-3) , b = -7, a=0$ valid

(3) $(x_1,x_2,x_3) = (1,-2,3) , b = -7, a=2$ valid

(4) $(x_1,x_2,x_3) = (-1,2,3) , b = 1, a=4$ valid

(5) $(x_1,x_2,x_3) = (-1,-2,-3) , b = 11$ invalid

the total event space is $21 \cdot (21- 1)$ (choice of select a $\cdot$ choice of selecting b given no-replacement)

hence, our answer is $\frac{4}{21 \cdot 20} = \boxed{\textbf{(C) }\frac{1}{105}}$

~luckuso

2024 AMC 12B (Problems • Answer Key • Resources)
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@@ Line 10: / Line 10: @@
 <math>-10 \leq a, b \leq 10</math>, each of <math>a,b</math> has <math>21</math> choices
-Applying Vieta,
+Applying Vieta's formulas,
 <math>x_1 \cdot x_2  \cdot x_3  = -6</math>

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

Revision as of 18:28, 16 November 2024

Problem 17

Solution 1

See also