3. The remainder of dividing a certain natural number $n$ by 22 is 7, and the remainder of dividing $n$ by 33 is 18. Find the remainder of dividing $n$ by 66.
Answer: 51.

Solution: According to the condition $n=22l+7, \quad l \in Z$ and $n=33m+18, m \in Z$. By equating these two expressions, we get $22l+7=33m+18, 2l=3m+1$. Since the left side of the equation is even, the right side must also be divisible by 2, so $m$ is an odd number, i.e., $m=2q+1, q \in Z$. Then $n=33(2q+1)+18=66q+51$, and the remainder of dividing $n$ by 70 is 51.