【Question 2】
$86 \times 87 \times 88 \times 89 \times 90 \times 91 \times 92 \div 7$ The remainder is $\qquad$.
【Analysis and Solution】
(Method One)
Since $91=7 \times 13$ is a multiple of 7;
Therefore, $86 \times 87 \times 88 \times 89 \times 90 \times 91 \times 92$ is also a multiple of 7; that is, the remainder of $86 \times 87 \times 88 \times 89 \times 90 \times 91 \times 92 \div 7$ is 0. (Method Two)
In any sequence of $n$ consecutive natural numbers, there must be one that is a multiple of $n$; in any sequence of 7 consecutive natural numbers, there must be one that is a multiple of 7. Therefore, among $86 \sim 92$, there must be one that is a multiple of 7;
Thus, $86 \times 87 \times 88 \times 89 \times 90 \times 91 \times 92$ is also a multiple of 7; that is, the remainder of $86 \times 87 \times 88 \times 89 \times 90 \times 91 \times 92 \div 7$ is 0.