Problem 1062

B2. Let $n$ be any natural number. The given expression is transformed as follows:

$$2^{2 n+3}+3^{n+2} \cdot 7^{n}=2^{3} \cdot 2^{2 n}+3^{2} \cdot 3^{n} \cdot 7^{n}=8 \cdot 4^{n}+9 \cdot 21^{n}=17 \cdot 4^{n}+9 \cdot\left(21^{n}-4^{n}\right)$$

Since the first term is a multiple of the number 17, it suffices to prove that $21^{n}-4^{n}$ is divisible by 17. This follows from the equality

$$21^{n}-4^{n}=(21-4)\left(21^{n-1}+21^{n-2} \cdot 4+\ldots+21 \cdot 4^{n-2}+4^{n-1}\right)$$

Reformulating the expression as $2^{3} \cdot 2^{2 n}+3^{2} \cdot 3^{n} \cdot 7^{n}$

3 points

Writing the expression as $8 \cdot 4^{n}+9 \cdot 21^{n}$ 4 points

Reformulating the expression as $17 \cdot 4^{n}+9 \cdot\left(21^{n}-4^{n}\right)$ 4 points

Factoring the expression $21^{n}-4^{n}$ 4 points

Logical conclusion about the divisibility of the expression by 17 5 points

2nd method. As in the first solution, we transform the expression to $8 \cdot 4^{n}+9 \cdot 21^{n}$. Since the remainder of the number 21 when divided by 17 is 4, the number $8 \cdot 4^{n}+9 \cdot 21^{n}$ has the same remainder when divided by 17 as the number $8 \cdot 4^{n}+9 \cdot 4^{n}=17 \cdot 4^{n}$. Since the latter is divisible by 17, the original number is also divisible by 17.

Reformulating the expression as $2^{3} \cdot 2^{2 n}+3^{2} \cdot 3^{n} \cdot 7^{n}$ 3 points

Writing the expression as $8 \cdot 4^{n}+9 \cdot 21^{n}$ 4 points

Logical conclusion that the expression $8 \cdot 4^{n}+9 \cdot 21^{n}$ has the same remainder when divided by 17 as the expression $17 \cdot 4^{n}$ 10 points

Conclusion that the expression is divisible by 17 3 points

## 62nd Mathematical Competition for High School Students in Slovenia

Selection Competition, March 15, 2018

Each mathematically correct and complete solution is worth $\mathbf{2 0}$ points, even if the solving process is different from the official solutions.

## Solutions for 2nd Year