Mock AIME 1 2007-2008 Problems/Problem 7

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Problem

Consider the following function $g(x)$ defined as \[(x^{2^{2008}-1}-1)g(x) = (x+1)(x^2+1)(x^4+1)\cdots (x^{2^{2007}}+1) - 1\] Find $g(2)$.

Solution

Multiply both sides by $x-1$; the right hand side collapses by the reverse of the difference of squares.

\begin{align*}(x-1)(x^{2^{2008}-1}-1)g(x) &= (x-1)(x+1)(x^2+1)(x^4+1)\cdots (x^{2^{2007}}+1) - (x-1)\\ &= (x^2-1) (x^2+1)(x^4+1)\cdots (x^{2^{2007}}+1) - (x-1)\\ &= \cdots\\ &= \left(x^{2^{2008}}-1\right) - (x-1) = x^{2^{2008}} - x \end{align*} Substituting $x = 2$, we have \[\left(2^{2^{2008}-1}-1\right) \cdot g(2) = 2^{2^{2008}}-2 = 2\left(2^{2^{2008}-1}-1\right)\] Dividing both sides by $2^{2^{2008}-1}$, we find $g(2) = \boxed{002}$.

See also

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by
Problem 6
Followed by
Problem 8
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