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 \(9\).