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

Revision as of 20:14, 3 December 2007

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

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

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

@@ Line 1: / Line 1: @@
 == Problem ==
-Let <math>f</math> be a function with the following properties:
+Let <math>f</math> be a [[function]] with the following properties:
 :<math>(i)\quad f(1) = 1</math>, and
@@ Line 10: / Line 10: @@
 == Solution ==
-<math>f(2^{100}) = f(2 \cdot 2^{99}) = 2^{99} \cdot f(2^{99}) = 2^{99} \cdot 2^{98} \cdot 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)}</math>.
+<math>f(2^{100}) = f(2 \times 2^{99}) = 2^{99} \times f(2^{99})</math> <math>= 2^{99} \cdot 2^{98} \times f(2^{98}) = \ldots</math> <math>= 2^{99}2^{98}\cdots 2^{1} \cdot 1 \cdot f(1)</math> <math>= 2^{99 + 98 + \ldots + 2 + 1}</math> <math>= 2^{\frac{99(100)}{2}} = 2^{4950}</math> <math>\Rightarrow \mathrm{(D)}</math>.
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Difference between revisions of "2004 AMC 12A Problems/Problem 17"

Revision as of 20:14, 3 December 2007

Problem

Solution

See also