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Difference between revisions of "2002 AMC 12P Problems/Problem 23"

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== Solution ==
 
== Solution ==
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If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
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Note that <math>2002 = 11 \cdot 13 \cdot 14</math>. With this observation, it becomes easy to note that <math>z = -14i</math> is a root of the given equation. However, it is not of the desired form in the problem, so we must factor the given expression to obtain the other 2 roots. From this point onwards, we assume that <math>z \neq -14i</math>.
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Expanding <math>z(z+i)(z+3i)=2002i</math>, we have <math>z^3 + 4iz^2 - 3z - 2002i = 0</math>. We may factor it as <math>(z + 14i)(z^2 - 10iz - 143) = 0</math>. Since
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<math>z + 14i \neq 0</math>, we must have <math>z^2 - 10iz - 143 = 0</math>. Therefore, <math>z = \frac {10i \pm \sqrt{-100 + 4(143)}}{2} = 5i \pm \sqrt {118}</math>.
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Since <math>a > 0</math>, we ignore the negative root. Therefore, <math>a = \boxed {\text{(A) } \sqrt {118}}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=22|num-a=24}}
 
{{AMC12 box|year=2002|ab=P|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:02, 1 July 2024

Problem

The equation $z(z+i)(z+3i)=2002i$ has a zero of the form $a+bi$, where $a$ and $b$ are positive real numbers. Find $a.$

$\text{(A) }\sqrt{118} \qquad \text{(B) }\sqrt{210} \qquad \text{(C) }2 \sqrt{210} \qquad \text{(D) }\sqrt{2002} \qquad \text{(E) }100 \sqrt{2}$

Solution

Note that $2002 = 11 \cdot 13 \cdot 14$. With this observation, it becomes easy to note that $z = -14i$ is a root of the given equation. However, it is not of the desired form in the problem, so we must factor the given expression to obtain the other 2 roots. From this point onwards, we assume that $z \neq -14i$.

Expanding $z(z+i)(z+3i)=2002i$, we have $z^3 + 4iz^2 - 3z - 2002i = 0$. We may factor it as $(z + 14i)(z^2 - 10iz - 143) = 0$. Since $z + 14i \neq 0$, we must have $z^2 - 10iz - 143 = 0$. Therefore, $z = \frac {10i \pm \sqrt{-100 + 4(143)}}{2} = 5i \pm \sqrt {118}$.

Since $a > 0$, we ignore the negative root. Therefore, $a = \boxed {\text{(A) } \sqrt {118}}$.

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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