59. What digit do the numbers

$$
6^{811}, 2^{1000}, 3^{999} 
$$

end with?

In 7-arithmetic, there are seven numbers:

$$
0,1,2,3,4,5,6 .
$$

Addition and multiplication in 7-arithmetic are defined by the following rules: to add two numbers, find their sum in the usual arithmetic and then take the remainder of this sum when divided by 7; to multiply two numbers, find their usual product and then take the remainder of this product when divided by 7.

Here are some examples:

$$
\begin{array}{ll}
3+5=1, & 5 \cdot 3=1 \\
4+6=3, & 3 \cdot 6=4 \\
3+4=0, & 2 \cdot 6=5
\end{array}
$$
59. To solve the problem, it is sufficient to compute the expressions $6^{811}$, $2^{1000}$, and $3^{999}$ in 10-arithmetic. In 10-arithmetic, the equality $6^{2}=6$ holds. Multiplying this equality successively by $6, 6^{2}, 6^{3}, \ldots$, we get

$$
6^{3}=6^{2}, 6^{4}=6^{3}, 6^{5}=6^{4}, \ldots
$$

From this, it follows that for any exponent $n$, $6^{n}=6$. In particular, $6^{811}=6$. To compute $2^{1000}$, note that $2^{4}=6$; therefore, $2^{1000}=2^{4 \cdot 250}=6^{250}=6$. Finally, from the equality $3^{4}=1$, we derive $3^{999}=3^{4 \cdot 249+3}=\left(3^{4}\right)^{249} \cdot 3^{3}=3^{3}=7$.