# 2004 AMC 12A Problems/Problem 17

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

## Solution 1 (Forward)

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*} f\left(2^{100}\right)&=2^{99+98+97+\cdots+3+2+1} \\ &=2^{99\cdot100/2} \\ &= \boxed{\textbf {(D)}\ 2^{4950}}. \end{align*} ~MRENTHUSIASM

## Solution 2 (Backward)

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{\textbf {(D)}\ 2^{4950}}. \end{align*} ~Azjps (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

## Video Solution

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