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

Revision as of 23:56, 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)

From (ii), note that $\begin{alignat*}{8} f(2) &= 1\cdot f(1) &&= 1, \\ f\left(2^2\right) &= 2\cdot f(2) &&= 2, \\ f\left(2^3\right) &= 2^2\cdot f\left(2^2\right) &&= 2^{2+1}, \\ f\left(2^4\right) &= 2^3\cdot f\left(2^3\right) &&= 2^{3+2+1}, \end{alignat*}$ and so on.

In general, we have $f\left(2^n\right)=2^{(n-1)+(n-2)+(n-3)+\cdots+3+2+1}$ for any positive integer $n.$

Therefore, the answer is $\begin{align*} 2^{100}&=2^{99+98+97+\cdots+3+2+1} \\ &=2^{99\cdot100/2} \\ &= \boxed{\text {(D)}\ 2^{4950}}. \end{align*}$ ~MRENTHUSIASM

Solution 2 (Backwards)

Applying (ii) repeatedly, we have $\begin{align*} f\left(2^{100}\right) &= 2^{99} \cdot f\left(2^{99}\right) \\ &= 2^{99} \cdot 2^{98} \cdot f\left(2^{98}\right) \\ &= 2^{99} \cdot 2^{98} \cdot 2^{97} \cdot f\left(2^{97}\right) \\ &= \cdots \\ &= 2^{99} \cdot 2^{98} \cdot 2^{97} \cdots 2^{3} \cdot 2^{2} \cdot 2^{1} \cdot 1 \cdot f(1) \\ &= 2^{99} \cdot 2^{98} \cdot 2^{97} \cdots 2^{3} \cdot 2^{2} \cdot 2^{1} \cdot 1 \cdot 1 \\ &=2^{99 + 98 + 97 + \cdots + 3 + 2 + 1} \\ &=2^{99\cdot100/2} \\ &= \boxed{\text {(D)}\ 2^{4950}}. \end{align*}$ ~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 33: / Line 33: @@
 == Solution 2 (Backwards) ==
-We have
+Applying (ii) repeatedly, we have
 <cmath>\begin{align*}
 f\left(2^{100}\right) &= 2^{99} \cdot f\left(2^{99}\right) \\

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