If $2^{200} \cdot 2^{203}+2^{163} \cdot 2^{241}+2^{126} \cdot 2^{277}=32^{n}$, what is the value of $n$ ?
Since $2^{a} \cdot 2^{b}=2^{a+b}$, then

$$
\begin{aligned}
32^{n} & =2^{200} \cdot 2^{203}+2^{163} \cdot 2^{241}+2^{126} \cdot 2^{277} \\
& =2^{200+203}+2^{163+241}+2^{126+277} \\
& =2^{403}+2^{404}+2^{403} \\
& =2^{403}+2^{403}+2^{404}
\end{aligned}
$$

Since $2^{c}+2^{c}=2\left(2^{c}\right)=2^{1} \cdot 2^{c}=2^{c+1}$, then

$$
\begin{aligned}
32^{n} & =2^{403+1}+2^{404} \\
& =2^{404}+2^{404} \\
& =2^{404+1} \\
& =2^{405}
\end{aligned}
$$

Since $\left(2^{d}\right)^{e}=2^{\text {de }}$, then $32^{n}=\left(2^{5}\right)^{n}=2^{5 n}$.

Since $32^{n}=2^{405}$, then $2^{5 n}=2^{405}$ which means that $5 n=405$ and so $n=81$.

ANSWER: $n=81$