Difference between revisions of "2004 AMC 12A Problems/Problem 17"
m |
MRENTHUSIASM (talk | contribs) m |
||
(7 intermediate revisions by 2 users not shown) | |||
Line 2: | Line 2: | ||
== Problem == | == Problem == | ||
− | Let <math>f</math> be a | + | Let <math>f</math> be a function with the following properties: |
− | + | (i) <math>f(1) = 1</math>, and | |
− | + | ||
+ | (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>\ | + | <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) | |
− | |||
==Video Solution== | ==Video Solution== | ||
− | https://youtu.be/ | + | https://youtu.be/qj5hBxYWalI |
== See also == | == See also == |
Latest revision as of 07:14, 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 be a function with the following properties:
(i) , and
(ii) for any positive integer .
What is the value of ?
Solution 1 (Forward)
From (ii), note that and so on.
In general, we have for any positive integer
Therefore, the answer is ~MRENTHUSIASM
Solution 2 (Backward)
Applying (ii) repeatedly, we have ~Azjps (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Video Solution
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.