Example 4 from the sequence of natural numbers
$$
1,2,3,4,5, \cdots
$$

cross out multiples of 3, multiples of 4, but among them, all multiples of 5 are retained without being crossed out. What is the 1995th number in the remaining sequence?
The least common multiple of $3, 4, 5$ is 60. In $1, 2, \cdots, 60$, the multiples of 3 are $20\left(=\frac{60}{3}\right)$, the multiples of 4 are $15\left(=\frac{60}{4}\right)$, the common multiples of 3 and 4 are $5\left(=\frac{60}{3 \times 4}\right)$, the common multiples of 3 and 5 are $4\left(=\frac{60}{3 \times 5}\right)$, the common multiples of 4 and 5 are $3\left(=\frac{60}{4 \times 5}\right)$, and the common multiples of $3, 4, 5$ are 1. Therefore, in $1, 2, \cdots, 60$, the number of remaining numbers is
$$
60-20-15+5+4+3-1=36
$$

Similarly, in
$$
60 m+1, 60 m+2, \cdots, 60 m+60 \quad(m \in \mathbf{N})
$$

there are 36 remaining numbers. Since
$$
1995=55 \times 36+15,
$$

and the 15th remaining number in $1, 2, \cdots, 60$ is 25, the 1995th remaining number in (6) is