Problem 955

## Solution.

Let's write the given expression as a product of two natural numbers.

$$\begin{aligned} 6^{2 n+2}-2^{n+3} \cdot 3^{n+2}+36 & =6^{2 n} \cdot 6^{2}-2^{n} \cdot 2^{3} \cdot 3^{n} \cdot 3^{2}+36 & & 1 \text{ point} \\ & =6^{2 n} \cdot 36-6^{n} \cdot 72+36 & & 1 \text{ point} \\ & =36\left(6^{2 n}-2 \cdot 6^{n}+1\right) & & 1 \text{ point} \\ & =36\left(6^{n}-1\right)^{2} & & 1 \text{ point} \end{aligned}$$

The obtained expression is clearly divisible by 36, and since $6^{n}-1$ is divisible by 5 for all natural numbers $n$, it follows that the given number is divisible by $36 \cdot 25=900$.

Note: The claim that $6^{n}-1$ is divisible by 5 follows from $6^{n}-1=(6-1)\left(6^{n-1}+6^{n-2}+\cdots+6+1\right)$ and does not need to be proven.