10. For what least natural \(m\) are the numbers of the form:
1) \(m^{3}+3^{m}\)
2) \(m^{2}+3^{m}\) 
divisible by 7?
10. Solution. Let \(r_{1}\) be the remainder of the division of \(m^{3}\) by 7, and \(r_{2}\) be the remainder of the division of \(3^{m}\) by 7. Clearly, the number \(m^{3}+3^{m}\) will be divisible by 7 when the sum of the remainders \(r_{1}+r_{2}\) equals 7. Let's construct the following table:

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline\(m\) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & \(\ldots\) \\
\hline\(r_{1}\) & 1 & 1 & 6 & 1 & 6 & 6 & \(\ldots\). & \(\ldots\) & \(\ldots\) & \(\ldots\) \\
\hline\(r_{2}\) & 3 & 2 & 6 & 4 & 5 & 1 & \(\ldots\) & \(\ldots\) & \(\ldots\). & \(\ldots\) \\
\hline\(r_{1}+r_{2}\) & 4 & 3 & 12 & 5 & 4 & 7 & \(\ldots\) & \(\ldots\) & \(\ldots\) & \(\ldots\) \\
\hline
\end{tabular}

From the table, it is clear that \(r_{1}+r_{2}=7\) when \(m=6\).