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Difference between revisions of "2004 AMC 12A Problems/Problem 17"

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{{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #17]] and [[2004 AMC 10A Problems/Problem 24|2004 AMC 10A 24]]}}
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== Problem ==
 
== Problem ==
 
Let <math>f</math> be a [[function]] with the following properties:
 
Let <math>f</math> be a [[function]] with the following properties:
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2004|ab=A|num-b=16|num-a=18}}
 
{{AMC12 box|year=2004|ab=A|num-b=16|num-a=18}}
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{{AMC10 box|year=2004|ab=A|num-b=23|num-a=25}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 15:12, 5 December 2007

The following problem is from both the 2004 AMC 12A #17 and 2004 AMC 10A 24, so both problems redirect to this page.

Problem

Let $f$ be a function with the following properties:

$(i)\quad f(1) = 1$, and
$(ii)\quad f(2n) = n\times f(n)$, for any positive integer $n$.

What is the value of $f(2^{100})$?

$\text {(A)}\ 1 \qquad \text {(B)}\ 2^{99} \qquad \text {(C)}\ 2^{100} \qquad \text {(D)}\ 2^{4950} \qquad \text {(E)}\ 2^{9999}$

Solution

$f(2^{100}) = f(2 \times 2^{99}) = 2^{99} \times f(2^{99})$ $= 2^{99} \cdot 2^{98} \times f(2^{98}) = \ldots$ $= 2^{99}2^{98}\cdots 2^{1} \cdot 1 \cdot f(1)$ $= 2^{99 + 98 + \ldots + 2 + 1}$ $= 2^{\frac{99(100)}{2}} = 2^{4950}$ $\Rightarrow \mathrm{(D)}$.

See also

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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
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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