Croatia_2018

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 $4$:

$k$	$2^k$	Last digit
1	2	2
2	4	4
3	8	8
4	16	6
5	32	2
6	64	4
7	128	8
8	256	6

So, the last digit of $2^k$ depends on k mod 4: - If $k\equiv 1(\mathrm{mod}4)$, last digit is $2$. - If $k\equiv 2(\mathrm{mod}4)$, last digit is $4$. - If $k\equiv 3(\mathrm{mod}4)$, last digit is $8$. - If $k\equiv 0(\mathrm{mod}4)$, last digit is $6$.

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

But more importantly, $2^{n+2}$ is divisible by $4$ for $n\ge 1$: - $2^{n+2}=4\cdot 2^n$, so $2^{n+2}\equiv 0(\mathrm{mod}4)$.

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

From the table above, if $k\equiv 0(\mathrm{mod}4)$, 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 $n$.