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\vskip 1cm{\LARGE\bf 
The Abundancy Index of Divisors of  \\
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Odd Perfect Numbers
}
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\large
Jose Arnaldo B. Dris \\
De La Salle University \\
Manila 1004 \\
The Philippines\\
\href{email:jabdris@yahoo.com.ph}{\tt jabdris@yahoo.com.ph} \\
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\begin{abstract}
If $N = {q^k}{n^2}$ is an odd perfect number, where $q$ is the Euler
prime, then we show that $\sigma(n) \le q^k$ is necessary and
sufficient for Sorli's conjecture that $k = \nu_{q}(N) = 1$ to 
hold.  It follows that, if $k = 1$ then the Euler prime $q$ is the
largest prime factor of $N$ and that $q > {10}^{500}$.  We also prove
that $q^k < \frac{2}{3}{n^2}$.
\end{abstract}


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