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

Revision as of 23:03, 9 July 2021

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) $f(1) = 1$ , and

(ii) $f(2n) = n \cdot f(n)$ 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 1 (Forwards)

Solution 2 (Backwards)

We have $\begin{align*} f\left(2^{100}\right) &= 2^{99} \cdot f\left(2^{99}\right) \\ &= 2^{99} \cdot 2^{98} \cdot f\left(2^{98}\right) \\ &= 2^{99} \cdot 2^{98} \cdot 2^{97} \cdot f\left(2^{97}\right) \\ &= \cdots \\ &= 2^{99} \cdot 2^{98} \cdot 2^{97} \cdots 2^{2} \cdot 2^{1} \cdot 1 \cdot f(1) \\ &= 2^{99} \cdot 2^{98} \cdot 2^{97} \cdots 2^{2} \cdot 2^{1} \cdot 1 \cdot 1 \\ &= 2^{99 + 98 + 97 + \cdots + 2 + 1} \\ &= 2^{\frac{100(99)}{2}} \\ &= \boxed{\text {(D)}\ 2^{4950}} \end{align*}$ ~Azjps (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Video Solution

https://youtu.be/qj5hBxYWalI

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 2: / Line 2: @@
 == 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
+(i) <math>f(1) = 1</math>, and
-:<math>(ii)\quad f(2n) = n\times f(n)</math>, for any positive integer <math>n</math>.
+(ii) <math>f(2n) = n \cdot f(n)</math> for any positive integer <math>n</math>.
 What is the value of <math>f(2^{100})</math>?
@@ Line 11: / Line 12: @@
 <math>\text {(A)}\ 1 \qquad \text {(B)}\ 2^{99} \qquad \text {(C)}\ 2^{100} \qquad \text {(D)}\ 2^{4950} \qquad \text {(E)}\ 2^{9999}</math>
-== Solution ==
+== Solution 1 (Forwards) ==
-<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>.
+== Solution 2 (Backwards) ==
+We have
+<cmath>\begin{align*}
+f\left(2^{100}\right) &= 2^{99} \cdot f\left(2^{99}\right) \\
+&= 2^{99} \cdot 2^{98} \cdot f\left(2^{98}\right) \\
+&= 2^{99} \cdot 2^{98} \cdot 2^{97} \cdot f\left(2^{97}\right) \\
+&= \cdots \\
+&= 2^{99} \cdot 2^{98} \cdot 2^{97} \cdots 2^{2} \cdot 2^{1} \cdot 1 \cdot f(1) \\
+&= 2^{99} \cdot 2^{98} \cdot 2^{97} \cdots 2^{2} \cdot 2^{1} \cdot 1 \cdot 1 \\
+&= 2^{99 + 98 + 97 + \cdots + 2 + 1} \\
+&= 2^{\frac{100(99)}{2}} \\
+&= \boxed{\text {(D)}\ 2^{4950}}
+\end{align*}</cmath>
+~Azjps (Fundamental Logic)
+~MRENTHUSIASM (Reconstruction)
 ==Video Solution==

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