Let $k, \alpha$ and $10 k-\alpha$ be positive integers. What is the remainder when the following number is divided by 11?

$$
8^{10 k+\alpha}+6^{10 k-\alpha}-7^{10 k-\alpha}-2^{10 k+\alpha}
$$
I. Solution. Let's form the remainders (less than 11, non-negative) of the powers of 2, then 6, 7, and 8 with exponents $1, 2, \ldots, 10$ when divided by 11 (see the table). In this, we use the fact that if in a product the factors are replaced by numbers that give the same remainder when divided by 11 as the original ones, then the new product also gives the same remainder when divided by 11 as the original. Indeed, if in the product $a \cdot b$, $a$ is replaced by $a_{1}$ and $b$ by $b_{1}$, where $a$ and $a_{1}$, and $b$ and $b_{1}$ give the same remainder when divided by 11, i.e., $a-a_{1}$ and $b-b_{1}$ are divisible by 11, then $a_{1} b_{1}$ and $a b$ also have the same remainder, i.e., $a b-a_{1} b_{1}$ is also divisible by 11:

$$
a b-a_{1} b_{1}=a b-a_{1} b+a_{1} b-a_{1} b_{1}=\left(a-a_{1}\right) b+a_{1}\left(b-b_{1}\right)
$$

because each term has one of its factors divisible by it.

| $n=$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | in case of |
| :--- | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| remainder of $2^{n}$ | 2 | 4 | 8 | 5 | 10 | 9 | 7 | 3 | 6 | 1 |  |
| remainder of $6^{n}$ | 6 | 3 | 7 | 9 | 10 | 5 | 8 | 4 | 2 | 1 |  |
| remainder of $7^{n}$ | 7 | 5 | 2 | 3 | 10 | 4 | 6 | 9 | 8 | 1 |  |
| remainder of $8^{n}$ | 8 | 9 | 6 | 4 | 10 | 3 | 2 | 5 | 7 | 1 |  |

According to this, the remainder of the 10th power of each of the four bases is 1. This makes it unnecessary to examine the further exponents, because the sequence of remainders for each base repeats every 10. Indeed, denoting any of the four bases by $a$,

$$
a^{10}=11 B+1
$$

so if the remainder of $a^{n}$ is $r_{2}$, i.e.,

$$
a^{n}=11 C+r_{2}
$$

then

$$
a^{n+10}=a^{n} \cdot a^{10}=\left(11 C+r_{2}\right)(11 B+1)=11 D+r_{2}
$$

where

$$
D=11 B C+B r_{2}+C
$$

and this proves our statement.

According to this, $\alpha=10 m$ (where $00)$, so the remainder of each power is 1, and the remainder of the expression is $1+1-1-1=0$. This means that instead of (1), it is enough to examine the

$$
\left(8^{\alpha}-7^{10-\alpha}\right)+\left(6^{10-\alpha}-2^{\alpha}\right)
$$

expression, where $0<\alpha<10$.

According to our table, the sequence of remainders of the base 6 is the reverse of the sequence of remainders of the base 2, so the difference $6^{10-\alpha}-2^{\alpha}$ always gives a remainder of 0 when divided by 11. The same holds for the sequences of remainders of the bases 7 and 8, and for the first bracketed expression in (2), so (2) and with it (1) always gives a remainder of 0, is divisible by 11.

II. Solution. Notice that the product of the bases of each of the minuend and subtrahend is 1 more than a multiple of 11:

$$
6 \cdot 2=11+1, \quad 8 \cdot 7=55+1
$$

We will show that this plays a crucial role in the correctness of the statement of the problem; generally, if $a \cdot b-1$ is divisible by 11, then

$$
D=a^{10 k+\alpha}-b^{10 k-\alpha}
$$

is divisible by 11, so $8^{10 k+\alpha}-7^{10 k-\alpha}$ and $6^{10 k+\alpha}-2^{10 k-\alpha}$ are also, and thus the expression in the problem is also divisible by 11.

Indeed,

$$
\begin{gathered}
D=a^{10 k+\alpha}-b^{10 k-\alpha}=a^{10 k+\alpha}-a^{\alpha} b^{10 k}+a^{\alpha} b^{10 k}-b^{10 k-\alpha}=a^{\alpha}\left[\left(a^{10}\right)^{k}-\left(b^{10}\right)^{k}\right]+ \\
b^{10 k-\alpha}\left[(a b)^{\alpha}-1\right]
\end{gathered}
$$

and here the first term is divisible by $c=a^{10}-b^{10}$, the second by $a b-1$. According to our condition, the latter is divisible by 11, so we only need to show that $c$ is divisible by 11.

From the condition, it follows that neither $a$ nor $b$ is divisible by 11, otherwise $a b$ would be divisible by 11, not $a b-1$. $c$ can be written as:

$$
c=\left(a^{10}-1\right)-\left(b^{10}-1\right)
$$

so it is enough to show that if an integer $d$ is not divisible by 11, then $d^{10}-1$ is divisible by it. We saw in the first solution that it is enough to show this for $d=1,2, \ldots, 10$. This is obvious for $d=1$, and we saw it for $d=2,6,7,8$; from the numbers $d=3,4,5,9,10$ we form the remainders of $d^{2}, d^{3}, d^{5}, d^{10}$, immediately replacing the powers with the remainders of their division by 11:

| $d=$ | 3 | 4 | 5 | 9 | 10 | in case of |
| :--- | ---: | ---: | ---: | ---: | ---: | :--- |
| remainder of $d^{2}$ | 9 | 5 | 3 | 4 | 1 |  |
| remainder of $d^{3}$ | 5 | 9 | 4 | 3 | 10 |  |
| remainder of $d^{5}$ | 1 | 1 | 1 | 1 | 10 |  |
| remainder of $d^{10}$ | 1 | 1 | 1 | 1 | 1 |  |

This completes our proof.

Remark. We could have avoided the numerical calculation of the last step by using Fermat's theorem, which states that if $p$ is a prime number and $a$ is not divisible by $p$, then $a^{p-1}-1$ is divisible by $p$.