3. In the known sequence $1,4,8,10,16,19,21,25,30,43$, the number of subarrays whose sum is divisible by 11 is $\qquad$ .
3. 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, resulting in the following sequence:
$$
1,4,-3,-1,5,-3,-1,3,-3,-1 .
$$

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 these, the equal ones are
$$
s_{1}=s_{4}=s_{10}=1, \quad s_{2}=s_{8}=5, \quad s_{3}=s_{7}=s_{9}=2 .
$$
$s_{4}-s_{1}, s_{10}-s_{1}, s_{10}-s_{4}, s_{8}-s_{2}, s_{7}-s_{3}, s_{9}-s_{3}, s_{9}-s_{7}$, a total of 7 pairs, are divisible by 11.