$9 \cdot 4$ can be divided by 3, and the digits of each number are limited to $1,2,3$ (1,2,3 may not all be used) of all different natural numbers less than 200000, how many are there?
(Shanghai, China High School Mathematics Competition, 1988)
[Solution 1] Since the necessary and sufficient condition for a natural number to be divisible by 3 is that the sum of its digits is divisible by 3, among the single-digit numbers, only 3 is divisible by 3.

For two-digit numbers, if the tens digit is 1, then the units digit is 2; if the tens digit is 2, then the units digit is 1; if the tens digit is 3, then the units digit is 3. Therefore, the two-digit numbers that satisfy the condition are only three: $12, 21, 33$.

For three-digit numbers, if the sum of the tens and hundreds digits is 2, then only 11 is possible, and the units digit is 1, i.e., the three-digit number is 111; if the sum of the tens and hundreds digits is 3, then only 12 and 21 are possible, and the units digit can only be 3, i.e., the three-digit numbers are 123, 213; if the sum of the tens and hundreds digits is 4, then 13, 31, 22 are possible, and the units digit can only be 2, i.e., the three-digit numbers are $132, 312, 222$; if the sum of the tens and hundreds digits is 5, then 23, 32 are possible, and the units digit can only be 1, i.e., the three-digit numbers are 231, 321; if the sum of the tens and hundreds digits is 6, then only 33 is possible, and the units digit can only be 3, i.e., the three-digit number is 333.
Therefore, there are $1+2+3+2+1=9=3^{2}$ three-digit numbers that satisfy the condition,
Similarly, there are $3^{3}$ four-digit numbers, and $3^{4}$ five-digit numbers that satisfy the condition.
For six-digit numbers, the first digit must be 1, and there are also $3^{4}$ such numbers.
Thus, the total number of natural numbers that satisfy the condition is
$$
1+3+3^{2}+3^{3}+2 \cdot 3^{4}=202 \text { (numbers). }
$$
[Solution 2] The $k$-digit numbers formed by the digits $1, 2, 3$ total $3^{k}$ numbers.
Arrange these $3^{k}$ numbers in ascending order:
$$
11 \cdots 11, \quad 11 \cdots 12, \quad 11 \cdots 13, \quad 11 \cdots 21, \quad 11 \cdots 22, \quad \cdots, \quad 33 \cdots 33 \text {. }
$$

We divide this sequence of numbers into groups of three from the beginning, so there are $\frac{3^{k}}{3}=3^{k-1}$ groups, because the three numbers in each group have different remainders when divided by 3, hence among these $3^{k}$ numbers, the numbers that leave a remainder of $0, 1, 2$ when divided by 3 each total $3^{k-1}$ numbers.

Thus, there is $3^{1-1}=1$ one-digit number divisible by 3. There are $3^{2-1}=3$ two-digit numbers divisible by 3, ..., and $3^{5-1}=3^{4}$ five-digit numbers divisible by 3. For six-digit numbers, the first digit must be 1, so the five digits following 1 must leave a remainder of 2 when divided by 3, and the number of such numbers is also $3^{5-1}=3^{4}$, thus the total is
$$
1+3+3^{2}+3^{3}+3^{4}+3^{4}=202 \text { (numbers). }
$$