Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2004 AMC 12A Problems/Problem 17"
m
 
(12 intermediate revisions by 4 users not shown)
Line 1: Line 1:
−
{{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #17]] and [[2004 AMC 10A Problems/Problem 24|2004 AMC 10A 24]]}}
+
{{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #17]] and [[2004 AMC 10A Problems/Problem 24|2004 AMC 10A #24]]}}
  
 
== Problem ==
 
== 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>?
 
What is the value of <math>f(2^{100})</math>?
  
−
<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>
+
<math>\textbf {(A)}\ 1 \qquad \textbf {(B)}\ 2^{99} \qquad \textbf {(C)}\ 2^{100} \qquad \textbf {(D)}\ 2^{4950} \qquad \textbf {(E)}\ 2^{9999}</math>
 +
 
 +
== Solution 1 (Forward) ==
 +
From (ii), note that
 +
<cmath>\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*}</cmath>
 +
and so on.
 +
 
 +
In general, we have <cmath>f\left(2^n\right)=2^{(n-1)+(n-2)+(n-3)+\cdots+3+2+1}</cmath> for any positive integer <math>n.</math>
 +
 
 +
Therefore, the answer is
 +
<cmath>\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*}</cmath>
 +
~MRENTHUSIASM
 +
 
 +
== Solution 2 (Backward) ==
 +
Applying (ii) repeatedly, 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^{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*}</cmath>
 +
~Azjps (Fundamental Logic)
 +
 
 +
~MRENTHUSIASM (Reconstruction)
  
−
== Solution ==
+
==Video Solution==
−
<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>.
+
https://youtu.be/qj5hBxYWalI
  
 
== See also ==
 
== See also ==
Line 19: Line 57:
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 08:14, 6 September 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 $f(2^{100})$?

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

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

https://youtu.be/qj5hBxYWalI

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_17&oldid=161680"