64. Determine the remainders of the number $2^{1000}$ when divided by 3, 5, 11, and 13.

Subtraction in $m$-arithmetic, like ordinary arithmetic, is defined as the inverse operation of addition: a number $x$ is called the difference of numbers $b$ and $a (x=b-a)$ if

$$
a+x=b
$$

For example, in 7-arithmetic,

$$
\begin{aligned}
& 2-5=4, \text { because } 5+4=2 \\
& 1-6=2, \text { because } 6+2=1
\end{aligned}
$$

The fastest way to compute the difference of two numbers in $m$-arithmetic is to find the difference in the usual sense and, if it is negative, increase it by $m$. For example, in 7-arithmetic,

$$
\begin{aligned}
& 1-5=-4=3 \\
& 2-3=-1=6
\end{aligned}
$$

Subtraction in $m$-arithmetic is always possible and leads to a unique answer. The use of the usual minus sign "一" for the $m$-arithmetic operation of subtraction is justified by the fact that if a formula from ordinary algebra is true and contains only the signs "+", "−", "·" and any number of parentheses, then such a formula remains true when interpreted in $m$-arithmetic. In particular, the formulas

$$
\begin{aligned}
& \left.-(-a)=a^{1}\right) \\
& -(a+b)=-a-b \\
& -(a-b)=-a+b \\
& a-1(-b)=a-b
\end{aligned}
$$

1) $-a$, as usual, denotes the number $0-a$. For example, in 8-arithmetic, $-3=5$.
as well as the formulas

$$
\begin{aligned}
a^{2}-b^{2} & =(a-b)(a+b) \\
(a-b)^{2} & =a^{2}-2 a b+b^{2} \\
(a-b)^{3} & =a^{3}-3 a^{2} b+3 a b^{2}-b^{3}
\end{aligned}
$$

and others.

$m$-arithmetic is applied to ordinary arithmetic to check operations of addition, subtraction, and multiplication. For this, the fact is used that a true equality from ordinary arithmetic turns into a true equality in $m$-arithmetic if each number is replaced by its remainder when divided by $m$. Let's check, for example, using 7-arithmetic the equality

$$
74218 \cdot 21363-81835=1585446299
$$

Replacing each number by its remainder when divided by 7, we get:

$$
4 \cdot 6-5=1
$$

This equality is not true in 7-arithmetic and, therefore, the original example contains an error. Most often, 9-arithmetic is used for checking. The reason is that any number $N$ and the sum of its digits give the same remainder when divided by 9, which greatly simplifies finding the remainders when divided by $9^{1}$ ).

Note that checking using $m$-arithmetic does not always allow the detection of errors. For example, if we were to check the incorrect equality (1) using 9-arithmetic, we would arrive at the correct 9-arithmetic equality

$$
4 \cdot 6-7=8
$$

To make the omission of an error less likely, one can perform the check simultaneously using several different $m$-arithmetics, for example, using 7-arithmetic and 9-arithmetic.

1) The numbers $10,10^{2}, 10^{3}, \ldots$ give a remainder of 1 when divided by 9. Therefore, the numbers

$$
N=a \cdot 10^{n}+b \cdot 10^{n-1}+\ldots+f \text { and } a+b+\ldots+f
$$

give the same remainder when divided by 9.
64. Answers: $1 ; 1 ; 1 ; 3$.