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

(New page: ==Problem== For each integer <math>n\geq 4</math>, let <math>a_n</math> denote the base-<math>n</math> number <math>0.\overline{133}_n</math>. The product <math>a_4a_5...a_{99}</math> can ...)
 
 
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==Problem==
 
==Problem==
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For each integer <math>n\geq 4</math>, let <math>a_n</math> denote the base-<math>n</math> number <math>0.\overline{133}_n</math>. The product <math>a_4a_5...a_{99}</math> can be expressed as <math>\frac {m}{n!}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is as small as possible. What is the value of <math>m</math>?
+
For each integer <math>n\geq 4</math>, let <math>a_n</math> denote the base-<math>n</math> number <math>0.\overline{133}_n</math>. The product <math>a_4a_5\cdots a_{99}</math> can be expressed as <math>\frac {m}{n!}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is as small as possible. What is <math>m</math>?
  
 
<math>\text {(A)} 98 \qquad \text {(B)} 101 \qquad \text {(C)} 132\qquad \text {(D)} 798\qquad \text {(E)}962</math>
 
<math>\text {(A)} 98 \qquad \text {(B)} 101 \qquad \text {(C)} 132\qquad \text {(D)} 798\qquad \text {(E)}962</math>
  
 
==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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<cmath>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</cmath>
+
Alternatively, we could have used the algebraic manipulation for repeating decimals,
  
−
<cmath>a_x*x^3=x^2+3x+3+a_x</cmath>
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<cmath>\begin{align*}
 +
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{align*}</cmath>
  
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<cmath>a_x(x^3-1)=x^2+3x+3</cmath>
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[[Telescoping]],
  
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<cmath>a_x=\frac{x^2+3x+3}{x^3-1}=\frac{(x+1)^3-1}{x(x^3-1)}</cmath>
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<cmath>\begin{align*}
 +
a_4a_5\cdots 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\cdots 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{align*}</cmath>
  
 +
Some factors cancel, (after all, <math>13 \cdot 37 \cdot 33 \cdot 6</math> isn't one of the answer choices)
  
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<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>
  
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<math>a_4a_5...a_{99}=\frac{999999}{4*5*6*\cdots*99*63}=\frac{13*37*33*6}{99!}</math>
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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>.
  
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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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==Solution 2==
  
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<math>99=33*3</math>
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Note that<cmath>0.\overline{133}_n = \frac{n^2+3n+3}{n^3-1},</cmath>by geometric series.
 +
Thus, we're aiming to find the value of<cmath>\prod_{k=4}^{99} \frac{k^2+3k+3}{k^3 - 1}.</cmath>Expanding the product out, this is equivalent to
 +
<cmath>\frac{4^2+3(4)+3}{4^3 - 1} \cdot \frac{5^2+3(5)+3}{5^3 - 1} \cdot \frac{6^2+3(6)+3}{6^3 - 1} \cdot ... \cdot \frac{99^2+3(99)+3}{99^3 - 1}.</cmath>Note that the numerator of the <math>a</math>th fraction and the denominator of the <math>a+1</math>th fraction for <math>1 \leq a \leq 95</math> cancel out to be <math>\frac{1}{a+3},</math> since<cmath>\frac{k^2 + 3k + 3}{(k+1)^3 - 1} = \frac{k^2 + 3k + 3}{k^3 + 3k^2 + 3k} = \frac{1}{k},</cmath>by the binomial theorem on the the denominator of the aforementioned. Since this forms a telescoping series, our product is now equivalent to<cmath>\frac{99^2 + 3(99) + 3}{4^3 - 1} \cdot \prod_{k=4}^{98} \frac{1}{k}, </cmath>which, after simplification gives <math>\frac{6}{98!} \cdot \frac{10101}{63} = \frac{962}{98!},</math> giving an answer of <math>\boxed{962}.</math>
  
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<math>\frac{13*37*33*6}{99!}=\frac{13*37*2}{98!}</math>
+
-fidgetboss_4000
  
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Since only the two goes into 98, n is at it's minimum.
+
==See Also==
 +
{{AMC12 box|year=2004|ab=A|num-b=24|after=Last Question}}
  
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<math>13*37*2=962 \Rightarrow \text {(E)}</math>
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[[Category:Intermediate Algebra Problems]]
−
 
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{{MAA Notice}}
−
==See Also==
 

Latest revision as of 13:50, 17 August 2020

Problem

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

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

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{align*} 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{align*}

Telescoping,

\begin{align*} a_4a_5\cdots 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\cdots 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{align*}

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 $13 \cdot 37 \cdot 2=962 \Rightarrow \text {(E)}$.

Solution 2

Note that\[0.\overline{133}_n = \frac{n^2+3n+3}{n^3-1},\]by geometric series. Thus, we're aiming to find the value of\[\prod_{k=4}^{99} \frac{k^2+3k+3}{k^3 - 1}.\]Expanding the product out, this is equivalent to \[\frac{4^2+3(4)+3}{4^3 - 1} \cdot \frac{5^2+3(5)+3}{5^3 - 1} \cdot \frac{6^2+3(6)+3}{6^3 - 1} \cdot ... \cdot \frac{99^2+3(99)+3}{99^3 - 1}.\]Note that the numerator of the $a$th fraction and the denominator of the $a+1$th fraction for $1 \leq a \leq 95$ cancel out to be $\frac{1}{a+3},$ since\[\frac{k^2 + 3k + 3}{(k+1)^3 - 1} = \frac{k^2 + 3k + 3}{k^3 + 3k^2 + 3k} = \frac{1}{k},\]by the binomial theorem on the the denominator of the aforementioned. Since this forms a telescoping series, our product is now equivalent to\[\frac{99^2 + 3(99) + 3}{4^3 - 1} \cdot \prod_{k=4}^{98} \frac{1}{k},\]which, after simplification gives $\frac{6}{98!} \cdot \frac{10101}{63} = \frac{962}{98!},$ giving an answer of $\boxed{962}.$

-fidgetboss_4000

See Also

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