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

(Created page with "== Problem == How many positive integers <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer? <math> \mathrm{(A) \ 0 } \qquad \mathrm{(B...")
 
m (Solution)
 
(3 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
How many positive [[integer]]s <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer?
+
The equation <math>z(z+i)(z+3i)=2002i</math> has a zero of the form <math>a+bi</math>, where <math>a</math> and <math>b</math> are positive real numbers. Find <math>a.</math>
  
<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 } </math>
+
<math>
 +
\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}
 +
</math>
  
 
== Solution ==
 
== 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 ==
 
== See also ==
{{AMC12 box|year=2000|num-b=6|num-a=8}}
+
{{AMC12 box|year=2002|ab=P|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:31, 10 March 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

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

See also

2002 AMC 12P (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png