3. In the known sequence $1,4,8,10,16,19,21,25,30,43$, how many groups of consecutive numbers have a sum that is divisible by 11?
3. Solution: There are 7 groups.

Since the problem involves divisibility by 11, the original sequence can be rewritten as $1,4,-3,-1,5,-3,-1$, $3,-3,-1$.
Let the sum of the first $n$ terms be $S_{n}$,
then $S_{1}=S_{4}=S_{10}=1, S_{2}=S_{8}=5, S_{3}=S_{7}=S_{9}=2$.
Thus, there are $C_{3}^{2}+C_{2}^{2}+C_{3}^{2}=7$ groups of numbers that can be divided by 11, which are $S_{4}-S_{1}, S_{10}-S_{1}, S_{10}-S_{4}, S_{8}-$ $S_{2}, S_{7}-S_{3}, S_{9}-S_{3}, S_{9}-S_{7}$. Therefore, there are 7 groups of consecutive numbers whose sum can be divided by 11.