1.5 Remove the multiples of 3 and 4 from the sequence of natural numbers $1,2,3,4, \cdots$, but retain all multiples of 5 (for example, 15 and 20 are not removed). After removing, write the remaining numbers in a sequence: $A_{1}=1, A_{2}=2, A_{3}=5, A_{4}=7, \cdots$ Find the value of $A_{1988}$.
(Advanced Math Test for Science Experimental Class Admission, 1988)
[Solution] First, estimate all natural numbers not greater than 100:
In $\{1,2,3, \cdots, 100\}$, there are $\left[\frac{100}{3}\right]=33$ multiples of 3, in $\{1,2,3, \cdots, 100\}$, there are $\left[\frac{100}{4}\right]=25$ multiples of 4, in $\{1,2,3, \cdots, 100\}$, there are $\left[\frac{100}{12}\right]=8$ multiples of 12, in $\{1,2,3, \cdots, 100\}$, there are $\left[\frac{100}{15}\right]=6$ multiples of 15, in $\{1,2,3, \cdots, 100\}$, there are $\left[\frac{100}{20}\right]=5$ multiples of 20, in $\{1,2,3, \cdots, 100\}$, there are $\left[\frac{100}{60}\right]=1$ multiples of 60. Therefore, by the principle of inclusion and exclusion, the sequence after the operation has $M_{100}=100-(33+25)+(8+6+5)-1=60$ (numbers).
This shows that, among the first 100 numbers, after the operation of crossing out numbers as described, 60 numbers remain, i.e., $\frac{3}{5}$ are retained. Therefore, to retain 1988 terms, at least $1988 \div \frac{3}{5} \approx 3314$ terms are needed, i.e., at least the numbers $1,2, \cdots, 3315$ need to be crossed out to possibly obtain $A_{1988}$.

In $1,2, \cdots, 3315$, there are $\left[\frac{3315}{3}\right]=1105$ multiples of 3, and $\left[\frac{3315}{4}\right]=828$ multiples of 4, and so on.
$$
\begin{array}{l}
\quad M_{3315}=3315-\left(\left[\frac{3315}{3}\right]+\left[\frac{3315}{4}\right]\right)+\left(\left[\frac{3315}{3 \cdot 4}\right]+\left[\frac{3315}{3 \cdot 5}\right]\right. \\
\left.\quad+\left[\frac{3315}{4 \cdot 5}\right]\right)-\left[\frac{3315}{3 \cdot 4 \cdot 5}\right] \\
=1989 .
\end{array}
$$

Since 3315 is divisible by 5, and 3314 is not divisible by 3 and 4, it has not been crossed out, thus
$$
A_{1988}=3314 .
$$