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2004 AMC 10A Problems/Problem 24: Difference between revisions

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==Problem==
==Problem==
Let <math>a_1,a_2,\cdots</math>, be a sequence with the following properties.
Let <math>a_1,a_2,\cdots</math>, be a [[sequence]] with the following properties.


     (i)  <math>a_1=1</math>, and
     (i)  <math>a_1=1</math>, and


     (ii)  <math>a_{2n}=n\cdot a_n</math> for any positive integer <math>n</math>.
     (ii)  <math>a_{2n}=n\cdot a_n</math> for any [[positive integer]] <math>n</math>.


What is the value of <math>a_{2^{100}}</math>?
What is the value of <math>a_{2^{100}}</math>?
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Note that
Note that


<math>a_2=2a_1</math>
<math>a_{2^n}=2^{n-1} a_{2^{n -1}}</math>


<math>a_{2^2}=2\cdot a_2=2\cdot1=2</math>
so that <math>a_{2^{100}} = 2^{99}\cdot a_{2^{99}} = 2^{99} \cdot 2^{98} \cdot a_{2^{98}} = \cdots = 2^{99}\cdot2^{98}\cdot\cdots\cdot2^1\cdot2^0 \cdot a_{2^0} = 2^{(1+99)\cdot99/2}=2^{4950}</math>


<math>a_{2^3}=4\cdot a_4=2^3\cdot2^{2+1}</math>
where in the last steps we use the [[exponent]] rule <math>b^x \cdot b^y = b^{x + y}</math> and the formula for the sum of an [[arithmetic series]].


<math>a_{2^8}=8\cdot a_8=2^3\cdot</math>


//The following is written by Dale Black -- Not Validated
==See also==
I think it should be <math>2^{4950}</math>
{{AMC10 box|year=2004|ab=A|num-b=23|num-a=25}}


<math>a_{2^{100}}=2^{99}\cdot2^{98}\cdot...\cdot2^1\cdot1=2^{(1+99)\cdot99/2}=2^{99\cdot50}=2^{4950}</math>
[[Category:Introductory Algebra Problems]]

Revision as of 15:58, 6 February 2007

Problem

Let $a_1,a_2,\cdots$, be a sequence with the following properties. 

    (i)  $a_1=1$, and

    (ii)  $a_{2n}=n\cdot a_n$ for any positive integer $n$.

What is the value of $a_{2^{100}}$?

$\mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2^{99} \qquad \mathrm{(C) \ } 2^{100} \qquad \mathrm{(D) \ } 2^{4050} \qquad \mathrm{(E) \ } 2^{9999}$

Solution

Note that

$a_{2^n}=2^{n-1} a_{2^{n -1}}$

so that $a_{2^{100}} = 2^{99}\cdot a_{2^{99}} = 2^{99} \cdot 2^{98} \cdot a_{2^{98}} = \cdots = 2^{99}\cdot2^{98}\cdot\cdots\cdot2^1\cdot2^0 \cdot a_{2^0} = 2^{(1+99)\cdot99/2}=2^{4950}$

where in the last steps we use the exponent rule $b^x \cdot b^y = b^{x + y}$ and the formula for the sum of an arithmetic series.


See also

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 10 Problems and Solutions
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