Theory:

How do we find whether a number is divisible by a certain number or not?
 
We use the traditional division method to find out the divisors. Instead, we can use a few quick hacks, and they are called divisibility tests.
 
As the name rightly suggests, divisibility rules or divisibility tests help us to quickly check if the given number is divisible by a certain number or not.

Some of the frequently used and basic divisibility tests are:
1. Divisibility test by \(10\)

2. Divisibility test by \(2\)

3. Divisibility test by \(5\)

4. Divisibility test by \(9\) and \(3\)
Divisibility test by \(10\)
Let us consider a few multiples of \(10\).
 
\(50\), \(100\), \(200\), \(300\),\(…\)
 
We can see that all these numbers end with \(0\), and it is the vital similarity between all multiples of \(10\).
 
Let us also observe a few other numbers, such as \(13\), \(89\), \(155\), etc. We find that all these numbers do not end with \(0\). Hence, these numbers are not divisible by \(10\).
 
Now let us consider the general form of a three-digit number '\(abc\)'.

\(100a + 10b + c\)
 
\(100a\) is a multiple of \(10\) as \(100\) itself is a multiple of \(10\). Any number multiplied by \(100\) automatically becomes a multiple of \(10\).
 
Similarly, \(10b\) is also a multiple of \(10\).
 
To make the three-digit number '\(abc\)' a multiple of \(10\), the digit '\(c\)' plays a crucial role.
 
The three-digit number '\(abc\)' will be a multiple of \(10\), only if '\(c\)' is \(0\).
 
In other words, a number is divisible by \(10\) only if it ends with a \(0\).
Divisibility by \(5\)
The numbers given below are all multiples of \(5\).
 
\(5\), \(10\), \(15\), \(20\), \(25\), \(30\), \(35\), \(40\),\(…\)

Now, when we observe the multiples carefully, we find that the numbers always end with a \(5\) or a \(0\).
 
We can conclude that if a number ends with a \(5\) or a \(0\), it is divisible by \(5\).

In general form, the three-digit number '\(abc\)' is \(100a + 10b + c\).
 
From the divisibility test by \(10\), we understood that both \(100a\) and \(10b\) are multiples of \(10\).
 
\(10 = 2 \times 5\)
 
Thus, \(100a\) and \(10b\) will also become a multiple of \(5\).
 
Hence, '\(c\)' should either be \(5\) or \(0\) to make the number '\(abc\)' a multiple of \(5\).
Divisibility by \(2\)
Let us look at a continuous range of even numbers given below.
 
\(2\), \(4\), \(6\), \(8\), \(10\), \(12\), \(14\), \(16\), \(18\), \(20\), \(22\), \(24\), \(26\), \(28\),\(…\)
 
After careful observation, we can understand that even numbers end only in \(2\), \(4\), \(6\), \(8\), and \(0\).
 
Consider any three-digit number '\(abc\)' and its general form \(100a + 10b + c\).
 
Here, \(100a\) and \(10b\) are multiples of \(2\) as \(100\) and \(10\) are multiples of \(2\) themselves.
 
Hence, '\(c\)' can hold any value \(2\), \(4\), \(6\), \(8\), and \(0\) to make '\(abc\)' divisible by \(2\).