jmc

We want to find the units digit of the quotient $$\frac{2^{1993}+3^{1993}}{5}.$$We list the final two digits of $2^n$ and $3^n$ in the next table. We also compute the units digit of the quotient whenever $2^n+3^n$ is divisible by $5.$

\begin{array}{|c|c|c|c|c|} \hline $n%%DISP_0%%amp;$2^n%%DISP_0%%amp;$3^n%%DISP_0%%amp;$2^n+3^n%%DISP_0%%amp;$\frac{2^n+3^n}5$\\ \hline 0&01&01&02&\\ 1&02&03&05&1\\ 2&04&09&13&\\ 3&08&27&35&7\\ 4&16&81&97&\\ 5&32&43&75&5\\ 6&64&29&93&\\ 7&28&87&15&3\\ 8&56&61&17&\\ 9&12&83&95&9\\ 10&24&49&73&\\ 11&48&47&95&9\\ 12&96&41&37&\\ 13&92&23&15&3\\ 14&84&69&53&\\ 15&68&07&75&5\\ 16&36&21&57&\\ 17&72&63&35&7\\ 18&44&89&33&\\ 19&88&67&55&1\\ 20&76&01&77&\\ 21&52&03&55&1\\ 22&04&09&13&\\ 23&08&27&35&7\\ 24&16&81&97&\\ 25&32&43&75&5\\ \hline \end{array}We notice that after the first pair, the sequence repeats every $20.$ Therefore $$2^{1993}+3^{1993}\equiv 2^{13}+3^{13}\equiv 15(\mathrm{mod}100).$$So, the units digit of the quotient $\frac{2^{1993}+3^{1993}}{5}$ is $\boxed{3}.$

(Note: "mod 100" essentially means "remainder when the number is divided by 100". So, $2^{1993}+3^{1993}\equiv 15(\mathrm{mod}100)$ means that $2^{1993}+3^{1993}$ is 15 more than a multiple of 100.)