The three positive integers $a, b$, and $c$ satisfy

$$
4^{a} \cdot 5^{b} \cdot 6^{c}=8^{8} \cdot 9^{9} \cdot 10^{10}
$$

Determine the value of $a+b+c$.
Solution

$$
\begin{aligned}
4^{a} \cdot 5^{b} \cdot 6^{c} & =8^{8} \cdot 9^{9} \cdot 10^{10} \\
2^{2 a+c} \cdot 5^{b} \cdot 3^{c} & =2^{24} \cdot 3^{18} \cdot 5^{10} \cdot 2^{10} \\
2^{2 a+c-34} \cdot 3^{c-18} \cdot 5^{b-10} & =1
\end{aligned}
$$

By the Fundamental Theorem of Arithmetic, all the exponents on the left side of the last equation are zero. Therefore, $c=18, 2a+c=34$, and $b=10$. Substituting the value of $c$ into the second equation, we find $a=8$. Thus, $a+b+c=18+8+10=36$.