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Difference between revisions of "2004 AMC 12A Problems/Problem 25"

(Too messy....)
(geo. series, {eqnarray*})
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==Solution==
 
==Solution==
 +
This is an infinite [[geometric series]] with common ratio <math>\frac{1}{x^3}</math> and initial term <math>x^{-1} + 3x^{-2} + 3x^{-3}</math>, so <math>a_x = \left(\frac{1}{x} + \frac{3}{x^2} + \frac{3}{x^3}\right)\left(\frac{1}{1-\frac{1}{x^3}}\right)</math> <math>= \frac{x^2 + 3x + 3}{x^3} \cdot \frac{x^3}{x^3 - 1}</math> <math>= \frac{x^2 + 3x + 3}{x^3 - 1}</math> <math>= \frac{(x+1)^3 - 1}{x(x^3 - 1)}</math>.
  
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<math>a_x = \frac{1}{x}+\frac{3}{x^2}+\frac{3}{x^3}+\frac{1}{x^4}+\frac{3}{x^5}+\frac{3}{x^6}+\cdots</math>
+
Alternatively, we could have used the algebraic manipulation for repeating decimals,
  
−
<math>a_x*x^3=x^2+3x+3+a_x</math>
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<cmath>\begin{eqnarray*}
 +
a_x &=& \frac{1}{x}+\frac{3}{x^2}+\frac{3}{x^3}+\frac{1}{x^4}+\frac{3}{x^5}+\frac{3}{x^6}+\cdots \\
 +
a_x \cdot x^3 &=& x^2+3x+3+a_x\\
 +
a_x(x^3-1) &=& x^2+3x+3\\
 +
a_x &=& \frac{x^2+3x+3}{x^3-1}=\frac{(x+1)^3-1}{x(x^3-1)}
 +
\end{eqnarray*}</cmath>
  
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<math>a_x(x^3-1)=x^2+3x+3</math>
+
[[Telescoping]],
  
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<math>a_x=\frac{x^2+3x+3}{x^3-1}=\frac{(x+1)^3-1}{x(x^3-1)}</math>
+
<cmath>\begin{eqnarray*}
 +
a_4a_5...a_{99}&=&\frac{(5^3-1)(6^3-1)\cdots (100^3-1)}{4 \cdot 5 \cdot 6 \cdot \cdots \cdot 99 \cdot (4^3-1)(5^3-1)\cdots(99^3-1)}\\
 +
a_4a_5...a_{99}&=&\frac{999999}{4 \cdot 5 \cdot 6 \cdot \cdots \cdot 99 \cdot 63}=\frac{13 \cdot 37 \cdot 33 \cdot 6}{99!}\end{eqnarray*}</cmath>
  
 +
Some factors cancel, (after all, <math>13 \cdot 37 \cdot 33 \cdot 6</math> isn't one of the answer choices)
  
−
<math>a_4a_5...a_{99}=\frac{(5^3-1)(6^3-1)\cdots (100^3-1)}{4*5*6*\cdots*99*(4^3-1)(5^3-1)\cdots(99^3-1)}</math>
+
<cmath>\frac{13 \cdot 37 \cdot 33 \cdot 6}{99!}=\frac{13 \cdot 37 \cdot 2}{98!}</cmath>
  
−
<math>a_4a_5...a_{99}=\frac{999999}{4*5*6*\cdots*99*63}=\frac{13*37*33*6}{99!}</math>
+
Since the only factor in the numerator that goes into <math>98</math> is <math>2</math>, <math>n</math> is minimized. Therefore the answer is <math>13 \cdot 37 \cdot 2=962 \Rightarrow \text {(E)}</math>.
−
 
 
−
Since <math>13*37*33*6</math> isn't one of the answer choices, we need to get rid of some stuff:
 
−
 
 
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<math>99=33*3</math>
 
−
 
 
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<math>\frac{13*37*33*6}{99!}=\frac{13*37*2}{98!}</math>
 
−
 
 
−
Since only the two goes into 98, n is at it's minimum.
 
−
 
 
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<math>13*37*2=962 \Rightarrow \text {(E)}</math>
 
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2004|ab=A|num-b=24|after=Final Question}}
 
{{AMC12 box|year=2004|ab=A|num-b=24|after=Final Question}}
 +
 +
[[Category:Introductory Algebra Problems]]

Revision as of 17:16, 4 December 2007

Problem

For each integer $n\geq 4$, let $a_n$ denote the base-$n$ number $0.\overline{133}_n$. The product $a_4a_5...a_{99}$ can be expressed as $\frac {m}{n!}$, where $m$ and $n$ are positive integers and $n$ is as small as possible. What is the value of $m$?

$\text {(A)} 98 \qquad \text {(B)} 101 \qquad \text {(C)} 132\qquad \text {(D)} 798\qquad \text {(E)}962$

Solution

This is an infinite geometric series with common ratio $\frac{1}{x^3}$ and initial term $x^{-1} + 3x^{-2} + 3x^{-3}$, so $a_x = \left(\frac{1}{x} + \frac{3}{x^2} + \frac{3}{x^3}\right)\left(\frac{1}{1-\frac{1}{x^3}}\right)$ $= \frac{x^2 + 3x + 3}{x^3} \cdot \frac{x^3}{x^3 - 1}$ $= \frac{x^2 + 3x + 3}{x^3 - 1}$ $= \frac{(x+1)^3 - 1}{x(x^3 - 1)}$.

Alternatively, we could have used the algebraic manipulation for repeating decimals,

\begin{eqnarray*} a_x &=& \frac{1}{x}+\frac{3}{x^2}+\frac{3}{x^3}+\frac{1}{x^4}+\frac{3}{x^5}+\frac{3}{x^6}+\cdots \\ a_x \cdot x^3 &=& x^2+3x+3+a_x\\ a_x(x^3-1) &=& x^2+3x+3\\ a_x &=& \frac{x^2+3x+3}{x^3-1}=\frac{(x+1)^3-1}{x(x^3-1)} \end{eqnarray*}

Telescoping,

\begin{eqnarray*} a_4a_5...a_{99}&=&\frac{(5^3-1)(6^3-1)\cdots (100^3-1)}{4 \cdot 5 \cdot 6 \cdot \cdots \cdot 99 \cdot (4^3-1)(5^3-1)\cdots(99^3-1)}\\ a_4a_5...a_{99}&=&\frac{999999}{4 \cdot 5 \cdot 6 \cdot \cdots \cdot 99 \cdot 63}=\frac{13 \cdot 37 \cdot 33 \cdot 6}{99!}\end{eqnarray*}

Some factors cancel, (after all, $13 \cdot 37 \cdot 33 \cdot 6$ isn't one of the answer choices)

\[\frac{13 \cdot 37 \cdot 33 \cdot 6}{99!}=\frac{13 \cdot 37 \cdot 2}{98!}\]

Since the only factor in the numerator that goes into $98$ is $2$, $n$ is minimized. Therefore the answer is $13 \cdot 37 \cdot 2=962 \Rightarrow \text {(E)}$.

See Also

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Final Question
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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