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

Revision as of 10:11, 16 October 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 $a_{2^{100}} = 2^{99}\cdot a_{2^{99}} = 2^{99} \cdot 2^{98} \cdot a_{2^{98}} = \cdots = 2^{99}\cdot2^{98}\cdot\cdots\cdot2^1\cdot2^0 \cdot a_{2^0}$

$= 2^{(1+99)\cdot99/2}=\boxed{2^{4950}}$

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.

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

@@ Line 15: / Line 15: @@
 <math>a_{2^n}=2^{n-1} a_{2^{n -1}}</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>
+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}</math>
+<math>= 2^{(1+99)\cdot99/2}=\boxed{2^{4950}}</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]].
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2004 AMC 10A Problems/Problem 24: Difference between revisions

Revision as of 10:11, 16 October 2007

Problem

Solution

See also