Solution
\[
\begin{aligned}
& 6^{m}+9^{n}=(5+1)^{m}+(10-1)^{n} \\
& =\left({ }^{m} C_{0} 5^{m}+\ldots+{ }^{m} C_{m-1} 5+{ }^{m} C_{m}\right)+\left({ }^{n} C_{0} 10^{n}-{ }^{n} C_{1} 10^{n-1}+\ldots+{ }^{n} C_{n-1} 10+{ }^{n} C_{n}(-1)^{n}\right] \\
& =(5 k+1)+\left(10 \lambda+(-1)^{n}\right) \\
& =5(k+2 \lambda)+1+(-1)^{n} \\
& m \in\{1,2,3,4, \ldots, 49,50\}
\end{aligned}
\]
n must be an odd no i.e. $\mathrm{n} \in\{1,3,5,7, \ldots, 49\}$;
\[
\begin{aligned}
& \mathrm{p}=50 \times 25=1250 \\
& \mathrm{~m}+\mathrm{n}=2^{2}=4 \\
& (\mathrm{~m}, \mathrm{n})=(1,3),(2,2),(3,1) \rightarrow 3 \\
& \mathrm{~m}+\mathrm{n}=3^{2}=9 \\
& (\mathrm{~m}, \mathrm{n})=(1,8),(2,7)(3,6), \ldots(8,1) \rightarrow 8 \\
& \mathrm{~m}+\mathrm{n}=5^{2}=25 \\
& (\mathrm{~m}, \mathrm{n})=(1,24)(2,23),(3,21) \ldots(24,1) \rightarrow 24 \\
& \mathrm{~m}+\mathrm{n}=7^{2}=49 \\
& (\mathrm{~m}, \mathrm{~m})=(1,48)(2,47),(3,46), \ldots(48,1) \rightarrow 48 \\
& (\mathrm{~m}+\mathrm{n})_{\max }=100 \\
& \mathrm{~m}+\mathrm{n} \neq 11^{2}=121
\end{aligned}
\]
\[
\begin{aligned}
& q=3+8+24+48=83 \\
& \therefore p+q=1250+83=1333 \text { Answer }(1333)
\end{aligned}
\]
.
