Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1999 AHSME Problems/Problem 13"

(Formatting)
 
(3 intermediate revisions by 2 users not shown)
Line 1: Line 1:
−
{{solution}}
 
−
 
 
==Problem==
 
==Problem==
  
−
Define a sequence of real numbers <math> a_1</math>, <math> a_2</math>, <math> a_3</math>, <math> \dots</math> by <math> a_1 = 1</math> and <math> a_{n\plus{}1}^3 = 99a_n^3</math> for all <math> n \geq 1</math>. Then <math> a_{100}</math> equals
+
Define a sequence of real numbers <math> a_1</math>, <math> a_2</math>, <math> a_3</math>, <math> \dots</math> by <math> a_1 = 1</math> and <math> a_{n + 1}^3 = 99a_n^3</math> for all <math> n \geq 1</math>. Then <math> a_{100}</math> equals
  
 
<math> \textbf{(A)}\ 33^{33} \qquad  
 
<math> \textbf{(A)}\ 33^{33} \qquad  
Line 10: Line 8:
 
\textbf{(D)}\ 99^{99} \qquad  
 
\textbf{(D)}\ 99^{99} \qquad  
 
\textbf{(E)}\ \text{none of these}</math>
 
\textbf{(E)}\ \text{none of these}</math>
 +
 +
==Solution==
 +
 +
We rearrange to get <math>\dfrac{a_{n+1}}{a_n} = \sqrt[3]{99}</math>. Thus we get <math>\dfrac{a_{n+1}}{a_n} = \sqrt[3]{99}</math>, <math>\dfrac{a_{n}}{a_{n-1}} = \sqrt[3]{99}</math>, and so on. Multiplying them all gives <math>\dfrac{a_{n+1}}{a_1} = (\sqrt[3]{99})^{n}</math>. Plugging in <math>n = 99</math> and <math>a_1 = 1</math>, <math>a_{100} = (\sqrt[3]{99})^{99} = 99^{33}</math>, so the answer is <math>\textbf{(C)}</math>.
  
 
==See Also==
 
==See Also==
  
 
{{AHSME box|year=1999|num-b=12|num-a=14}}
 
{{AHSME box|year=1999|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Latest revision as of 13:34, 5 July 2013

Problem

Define a sequence of real numbers $a_1$, $a_2$, $a_3$, $\dots$ by $a_1 = 1$ and $a_{n + 1}^3 = 99a_n^3$ for all $n \geq 1$. Then $a_{100}$ equals

$\textbf{(A)}\ 33^{33} \qquad \textbf{(B)}\ 33^{99} \qquad \textbf{(C)}\ 99^{33} \qquad \textbf{(D)}\ 99^{99} \qquad \textbf{(E)}\ \text{none of these}$

Solution

We rearrange to get $\dfrac{a_{n+1}}{a_n} = \sqrt[3]{99}$. Thus we get $\dfrac{a_{n+1}}{a_n} = \sqrt[3]{99}$, $\dfrac{a_{n}}{a_{n-1}} = \sqrt[3]{99}$, and so on. Multiplying them all gives $\dfrac{a_{n+1}}{a_1} = (\sqrt[3]{99})^{n}$. Plugging in $n = 99$ and $a_1 = 1$, $a_{100} = (\sqrt[3]{99})^{99} = 99^{33}$, so the answer is $\textbf{(C)}$.

See Also

1999 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1999_AHSME_Problems/Problem_13&oldid=56574"