3. In the known sequence $1,4,8,10,16,19,21,25,30,43$, the number of arrays where the sum of consecutive numbers is divisible by 11 is $\qquad$.
7
3.【Analysis and Solution】Since the problem is about divisibility by 11, we can first subtract multiples of 11 from each term to make the numbers smaller and easier to handle, thus obtaining the following sequence:
$$
1,4,-3,-1,5,-3,-1,3,-3,-1 \text {. }
$$

Let $S_{n}$ be the sum of the first $n$ terms, then
$$
\begin{array}{l}
S_{1}=1, S_{2}=5, S_{3}=2, S_{4}=1, S_{5}=6, \\
S_{6}=3, S_{7}=2, S_{8}=5, S_{9}=2, S_{10}=1 .
\end{array}
$$

Among them, the equal ones are
$$
\begin{array}{l}
S_{1}=S_{4}=S_{10}=1, S_{2}=S_{8}=5, S_{3}=S_{7}=S_{9}=2 . \\
S_{4}-S_{1}, S_{10}-S_{4}, S_{8}-S_{2}, S_{7}-S_{3}, S_{9}-S_{3}, S_{9}-S_{7}, S_{10}-S_{1} .
\end{array}
$$

There are 7 groups that can be divided by 11.