### Theory:

Before this, we discussed divisibility tests by $$10$$, $$5$$, and $$2$$.

What do you think is common among all three divisibility tests?

These three divisibility tests solely depended only on one-digit, that is, the last digit.

Hence, these three divisibility tests followed a similar procedure.

Let us now look at a comparatively, unique divisibility test.
Divisibility by $$3$$ or $$9$$
Consider the number $$4239$$

The expanded form is $$(4 \times 1000) + (2 \times 100) + (3 \times 10) + 9$$.

$$= [4 \times (999 + 1)] + [2 \times (99 + 1) + [3 \times (9 + 1)] + 9$$

$$= (4 \times 999) + (4 \times 1) + (2 \times 99) + (2 \times 1) + (3 \times 9) + (3 \times 1) + (9 \times 1)$$

$$= (4 \times 999) + (2 \times 99) + (3 \times 9) + (4 + 2 + 3 + 9)$$

Now, $$4 + 2 + 3 + 9 = 18$$.

$$18$$ is both divisible by $$3$$ and by $$9$$. Therefore, $$4239$$ is also divisible by $$3$$ or $$9$$.

Similarly, consider the number $$2148$$

The expanded form is $$(2 \times 1000) + (1 \times 100 + (4 \times 10) + 8$$.

$$= [2 \times (999 + 1)] + [1 \times (99 + 1) + [4 \times (9 + 1)] + 8$$

$$= (2 \times 999) + (2 \times 1) + (1 \times 99) + (1 \times 1) + (4 \times 9) + (4 \times 1) + (8 \times 1)$$

$$= (2 \times 999) + (1 \times 99) + (4 \times 9) + (2 + 1 + 4 + 8)$$

Now, $$2 + 1 + 4 + 8 = 15$$.

$$15$$ is divisible by $$3$$ but not by $$9$$. Therefore, $$2148$$ is also divisible by $$3$$ but not by $$9$$.