Difference between revisions of "2004 AMC 12A Problems/Problem 17"
MRENTHUSIASM (talk | contribs) m (→Solution 2 (Backwards)) |
MRENTHUSIASM (talk | contribs) m (→Solution 1 (Forwards)) |
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Line 26: | Line 26: | ||
Therefore, the answer is | Therefore, the answer is | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | 2^{100}&=2^{99+98+97+\cdots+3+2+1} \\ | + | f\left(2^{100}\right)&=2^{99+98+97+\cdots+3+2+1} \\ |
&=2^{99\cdot100/2} \\ | &=2^{99\cdot100/2} \\ | ||
&= \boxed{\text {(D)}\ 2^{4950}}. | &= \boxed{\text {(D)}\ 2^{4950}}. |
Revision as of 00:00, 10 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 be a function with the following properties:
(i) , and
(ii) for any positive integer .
What is the value of ?
Solution 1 (Forwards)
From (ii), note that and so on.
In general, we have for any positive integer
Therefore, the answer is ~MRENTHUSIASM
Solution 2 (Backwards)
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.