Difference between revisions of "2004 AMC 12A Problems/Problem 25"
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==Problem== | ==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\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 | + | 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> | ||
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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>. | 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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | -fidgetboss_4000 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2004|ab=A|num-b=24|after=Last Question}} | {{AMC12 box|year=2004|ab=A|num-b=24|after=Last Question}} | ||
− | [[Category: | + | [[Category:Intermediate Algebra Problems]] |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:50, 17 August 2020
Contents
Problem
For each integer , let
denote the base-
number
. The product
can be expressed as
, where
and
are positive integers and
is as small as possible. What is
?
Solution
This is an infinite geometric series with common ratio and initial term
, so
.
Alternatively, we could have used the algebraic manipulation for repeating decimals,
Some factors cancel, (after all, isn't one of the answer choices)
Since the only factor in the numerator that goes into is
,
is minimized. Therefore the answer is
.
Solution 2
Note thatby geometric series.
Thus, we're aiming to find the value of
Expanding the product out, this is equivalent to
Note that the numerator of the
th fraction and the denominator of the
th fraction for
cancel out to be
since
by the binomial theorem on the the denominator of the aforementioned. Since this forms a telescoping series, our product is now equivalent to
which, after simplification gives
giving an answer of
-fidgetboss_4000
See Also
2004 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last 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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