3. From the sequence of natural numbers $1, 2, 3, 4, 5, \cdots$, successively strike out multiples of 4, multiples of 7, but retain all multiples of 5 that are not multiples of 4 or 7. What is the 1995th number in the remaining sequence?
3. It can be solved as in Example 4 of Section 1.11.

Another solution: The least common multiple of $4,5,7$ is 140. By the Chinese Remainder Theorem (Sunzi's Theorem), numbers from 1 to 140 can be uniquely represented in the form $(a, b, c)$, where $a, b, c$ are the remainders when the number is divided by $4,5,7$ respectively. The retained numbers are of the form $(a, 0, c)$ and $(a, b, c), b \neq 0$. For the former, $a \in\{0,1,2,3\}, c \in\{0,1,2,3,4,5,6\}$, totaling $4 \times 7=$ 28 numbers; for the latter, $a \in\{1,2,3\}, b \in\{1,2,3,4\}, c \in\{1,2,3,4,5,6\}$, totaling $3 \times 4 \times 6=72$ numbers. Therefore, among the numbers from 1 to 140, a total of $28+72=100$ numbers are retained, with the largest five numbers being $140,139,138,137,135$. In the first $140 \times 20=2800$ natural numbers, $100 \times 20=2000$ numbers are retained. Therefore, the 1995th number is $2800-(140-$ $135)=2795$.