How many solutions has the equation:
$\{20 \cdot \{ 13 \cdot\{20 \cdot\{ 13\cdot x\}\} \}\}=x^{2013}$ -?
Here $\{z\}=z-[z]$, where $[z]=m \in \mathbb{Z}, \ m \le z < m+1$.
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Not, I trust, an ongoing contest? – Gerry Myerson Mar 19 '13 at 23:05
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No, of course . – Gordon Mar 19 '13 at 23:09
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Good. What contest, then? – Gerry Myerson Mar 20 '13 at 00:12
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$20\cdot 13\cdot 20\cdot 13$ nonnegative solutions. $x^{2013}$ is in $[0,1]$ for $x\in[0,1]$. For nonnegative $x$, the LHS is a collection of affine pieces with slopes $20\cdot 13\cdot 20\cdot 13$. Each piece covers $y$-axis values between $0$ and $1$. In $[0,1]$ there are $20\cdot 13\cdot 20\cdot 13$ such pieces and $x^{2013}$ intersects with each of them once.
Sander Heinsalu
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Interesting, something about the slope of $x^{2013}$ perhaps, e.g. intersects some affine piece more than once. If this is the case, then probably near $1$. This thing can be numerically calculated in e.g. Mathematica, but I don't want to spend the time now. – Sander Heinsalu Mar 20 '13 at 21:56