, 2022

We will prove that the smallest sum is equal to $8$. One case when it is achieved is for $n = 1$.

Suppose that for some $n > 1$ it is possible to obtain a smaller sum than $8$. Observing the last digit of the number $5^n + 6^n + 2022^n$, we can easily conclude that
$$5^n + 6^n + 2022^n \equiv \begin{cases} 7 \pmod{10}, & \text{if } n \equiv 0 \pmod{4} \\ 3 \pmod{10}, & \text{if } n \equiv 1 \pmod{4} \\ 5 \pmod{10}, & \text{if } n \equiv 2 \pmod{4} \\ 9 \pmod{10}, & \text{if } n \equiv 3 \pmod{4} \end{cases}$$
It follows that $n \equiv 1, 2 \pmod{4}$. We now consider these two cases.

Case 1: If $n = 4k + 1$, then
$$5^n + 6^n + 2022^n \equiv \begin{cases} 5 \pmod{9}, & \text{if } k \equiv 0 \pmod{3} \\ 2 \pmod{9}, & \text{if } k \equiv 1 \pmod{3} \\ 8 \pmod{9}, & \text{if } k \equiv 2 \pmod{3} \end{cases}$$
From here, due to the last digit being equal to $3$, it is only possible for the sum of the digits to be equal to $5$. Since $5^n + 6^n + 2022^n \equiv 1 \pmod{4}$, the penultimate digit must be equal to $1$, and all other digits (except the first) are equal to $0$. So the last digits of $5^n + 6^n + 2022^n$ are $0013$ and therefore $5^n + 6^n + 2022^n \equiv 13 \pmod{16}$. On the other hand, since $n > 1$ and $n \equiv 1 \pmod{4}$ we have $5^n + 6^n + 2022^n \equiv 5 \cdot 5^{4k} \equiv 5 \pmod{16}$, a contradiction.

Case 2: If $n = 4k + 2$, then
$$5^n + 6^n + 2022^n \equiv \begin{cases} 7 \pmod{9}, & \text{if } k \equiv 0 \pmod{3} \\ 1 \pmod{9}, & \text{if } k \equiv 1 \pmod{3} \\ 4 \pmod{9}, & \text{if } k \equiv 2 \pmod{3} \end{cases}$$
Due to the last digit, the only possibility is $k \equiv 0 \pmod{3}$ so $n = 12s + 2$ for some $s \in \mathbb{N}_0$. Now the sum of the digits is $7$, and the last digit is $5$. Let's use the divisibility criterion with $37$. (A number when divided by $37$ gives the same remainder as the sum of its three-digit blocks that are formed from right to left.) Since $6^n + 5^n + 2022^n$ must be equal to one of $20\cdots05$ or $10\cdots010\cdots05$, according to the above criterion $6^n+5^n+2022^n \equiv 205, 25, 7, 115, 106, 16 \pmod{37}$. I.e. congruent to $20, 25, 7, 4, 32, 16 \pmod{37}$.

On the other hand, since it is easy to check that $6^{12} \equiv 1 \pmod{37}$, $5^{12} \equiv 10 \pmod{37}$ and $2022^{12} \equiv 24^{12} \equiv 10 \pmod{37}$ we have that
$$5^n + 6^n + 2022^n \equiv 25 \cdot 10^2 + 36 + 24^2 \cdot 10^2 \equiv 36 + 9 \cdot 10^s \equiv \begin{cases} 8 \pmod{37}, & \text{if } s \equiv 0 \pmod{3} \\ 15 \pmod{37}, & \text{if } s \equiv 1 \pmod{3} \\ 11 \pmod{37}, & \text{if } s \equiv 2 \pmod{3} \end{cases}$$
This is a contradiction.

Let $A_n = 5^n + 6^n + 2022^n$. We have $A_n \equiv 1 \pmod{4}$. For $n \equiv 1, 2, \dots, 5 \pmod{5}$ we have $6^n \equiv 6, 11, 16, 21, 1 \pmod{25}$ and for $n \equiv 1, 2, \dots, 20 \pmod{20}$ we have
$$2022^n \equiv -3, 9, -2, 6, 7, 4, 13, 11, 17, -1, 3, -9, 2, -6, -7, -4, -13, -11, -17, 1 \pmod{25}$$
Thus for $n > 1$ and $n \equiv 1, 2, \dots, 20 \pmod{20}$ we have
$$A_n \equiv 3, 20, 14, 27, 8, 10, 24, 27, 38, 0, 9, 2, 18, 15, -6, 2, -2, 5, 4, 2 \pmod{25}$$
So for $n > 1$ and $n \equiv 1, 2, \dots, 20 \pmod{20}$ we have
$$A_n \equiv 53, 45, 89, 77, 33, 85, 49, 77, 13, 25, 9, 77, 93, 65, 69, 77, 73, 5, 29, 77 \pmod{100}$$
The only possibilities for the sum of the digits of $A_n$ to be less than $8$ are $n \equiv 5, 9, 18 \pmod{20}$ where the last two digits of $A_n$ are $33, 13, 05$ respectively. (The case $n \equiv 10 \pmod{20}$ is rejected since then we must have $A_n = 25$ which is impossible.)

Now for $n > 1$ and $n \equiv 1, 2, \dots, 6 \pmod{6}$ we have $A_n \equiv 5, 7, 8, 4, 2, 1 \pmod{9}$. Looking modulo $60$, the only possibilities are $n \equiv 5, 9, 18, 25, 29, 38, 45, 49, 58 \pmod{60}$ and in those cases the last two digits of $A_n$ are $33, 13, 05, 33, 13, 05, 33, 13, 05$ respectively and the sum of digits of $A_n$ are $2, 8, 1, 5, 2, 7, 8, 5, 4$ $\pmod{9}$ respectively.

So the only possibilities are $n \equiv 38, 49 \pmod{60}$ where the last two digits of $A_n$ are $05$ and $13$ respectively, and the sum of digits of $A_n$ are $7$ and $5$ respectively.

The only possibilities are therefore for $A_n$ to be equal to a number of the form $20\dots05$ or $10\dots010\dots05$ or $10\dots013$. In particular, using the alternating sum of digits criterion for divisibility by $11$ we have that the first number is congruent to $3, 7 \pmod{11}$, the second congruent to $3, 5, 7 \pmod{11}$ and the third congruent to $1, 3 \pmod{11}$.

We now look at $A_n \pmod{11}$. It can be easily checked that for $n \equiv 8 \pmod{10}$ we have
$$A_n \equiv 4 + 4 + 3 \equiv 0 \pmod{11}$$
and for $n \equiv 9 \pmod{10}$ we have
$$A_n \equiv 9 + 2 + 5 \equiv 5 \pmod{11}$$
So all three cases lead to a contradiction.