Croatia competition problems

Let us consider $2^{2^{n+2}} + 4$ modulo $10$.

First, note that $2^{2^{n+2}}$ is a very large power of $2$. Let's analyze the last digit of $2^k$ for large $k$.

The last digit of powers of $2$ cycles every $4$:

Now, $2^{n+2}$ is always even for $n \geq 1$ (since $n+2 \geq 3$), so $2^{n+2} \geq 8$.

But more importantly, $2^{n+2}$ is divisible by $4$ for $n \geq 1$:
- $2^{n+2} = 4 \cdot 2^{n}$, so $2^{n+2} \equiv 0 \pmod{4}$.

Therefore, $2^{2^{n+2}}$ is a power of $2$ where the exponent is divisible by $4$.

From the table above, if $k \equiv 0 \pmod{4}$, the last digit of $2^k$ is $6$.

Thus, $2^{2^{n+2}}$ ends with $6$ for any positive integer $n$.

Therefore,
$$2^{2^{n+2}} + 4$$
has last digit $6 + 4 = 10$, so ends with $0$.

Thus, $2^{2^{n+2}} + 4$ is divisible by $10$ for any positive integer $n$.