### 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, \(4 + 2 + 3 + 9 = 18\).

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, \(2 + 1 + 4 + 8 = 15\).

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 \(9\).