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

Latest revision as of 21:31, 10 March 2024

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 $a.$

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

According to Wolfram-Alpha, the answer is $\boxed {\text{(A) }\sqrt{118}}$ .

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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 == Solution ==
-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>.
+According to Wolfram-Alpha, the answer is <math>\boxed {\text{(A) }\sqrt{118}}</math>.
 == See also ==
 {{AMC12 box|year=2002|ab=P|num-b=22|num-a=24}}
 {{MAA Notice}}