Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 7"
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&= \left(x^{2^{2008}}-1\right) - (x-1) = x^{2^{2008}} - x \end{align*} </cmath> | &= \left(x^{2^{2008}}-1\right) - (x-1) = x^{2^{2008}} - x \end{align*} </cmath> | ||
Substituting <math>x = 2</math>, we have | Substituting <math>x = 2</math>, we have | ||
| − | <cmath>2^{2^{2008}-1} \cdot g(2) = 2^{2^{2008}}-2 = 2\left(2^{2^{2008}-1}-1\right)</cmath> | + | <cmath>\left(2^{2^{2008}-1}-1\right) \cdot g(2) = 2^{2^{2008}}-2 = 2\left(2^{2^{2008}-1}-1\right)</cmath> |
Dividing both sides by <math>2^{2^{2008}-1}</math>, we find <math>g(2) = \boxed{002}</math>. | Dividing both sides by <math>2^{2^{2008}-1}</math>, we find <math>g(2) = \boxed{002}</math>. | ||
Revision as of 20:14, 2 April 2008
Problem
Consider the following function 
defined as
![\[(x^{2^{2008}-1}-1)g(x) = (x+1)(x^2+1)(x^4+1)\cdots (x^{2^{2007}}+1) - 1\]](http://latex.artofproblemsolving.com/d/a/3/da37d7e641f65e65700d6574e6546587720b0199.png)
Find 
.
Solution
Multiply both sides by 
; the right hand side collapses by the reverse of the difference of squares.
Substituting 
, we have
![\[\left(2^{2^{2008}-1}-1\right) \cdot g(2) = 2^{2^{2008}}-2 = 2\left(2^{2^{2008}-1}-1\right)\]](http://latex.artofproblemsolving.com/8/9/4/8948f9f2a7545f68353ee2b2216e8c02e0074ede.png)
Dividing both sides by 
, we find 
.
See also
| Mock AIME 1 2007-2008 (Problems, Source) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
