Theory:

Divisibility by \(5\): If a digit in the ones place of a number is \(5\) or \(0\), then it is divisible by \(5\).
Example:
1. Let us take the numbers \(95\) and \(680\).
 
Rule for \(5\): Unit digit of a number is either \(5\) or \(0\).
 
The numbers \(95\) and \(680\) ends in \(5\) and \(0\).
 
Therefore, \(95\) and \(680\) are divisible by \(5\).
 
2. Let us take the numbers \(462\) and \(327\).
 
Rule for \(5\): Unit digit of a number is either \(5\) or \(0\).
 
The numbers \(462\) and \(327\) ends in \(2\) and \(7\).
 
Therefore, \(462\) and \(327\) are not divisible by \(5\).
Divisibility by \(6\): If a number is divisible by \(2\) and \(3\), then that number is divisible by \(6\).
Example:
1. Let us take the number \(246\).
 
Rule for \(6\): Number is divisible by \(2\) and \(3\).
 
It is divisible by \(2\) as it ends with \(6\).
 
Now, \(2+4+6=12\). \(12\) is divisible by \(3\), so \(246\) is divisible by \(3\) aslo.
 
This shows that \(246\) is divisible by \(2\) and \(3\).
 
Therefore, \(246\) is divisible by \(6\).
 
2. Let us take the number \(154\).
 
Rule for \(6\): Number is divisible by \(2\) and \(3\).
 
It is divisible by \(2\) as it ends with \(4\).
 
Now, \(1+5+4=10\). \(10\) is not divisible by \(3\).
 
Thus, \(154\) is not divisible by \(3\).
 
Therefore, \(154\) is not divisible by \(6\).
Divisibility by \(7\): Double the last digit of the number and then subtract it from the remaining number if the number formed is divisible by \(7\), then the number is divisible by \(7\).
Example:
1. Let us take the number \(2548\).
 
Rule for \(7\): Double the last digit and subtract it from the remaining number and check the result is divisible by \(7\).
 
Doubling the last digit \(8\) results as \(2\times 8 =16\).
 
Subtract it from the remaining number \( = 254-16 = 238\).
 
Now the number \(238\div 7 = 34\).
 
Therefore, \(2548\) is divisible by \(7\).
 
2. Let us take the number \(810\).
 
Rule for \(7\): Double the last digit and subtract it from the remaining number and check the result is divisible by \(7\).
 
Doubling the last digit \(0) results as \(2\times 0 = 0\). 
 
Subtract it from the remaining number \( = 81- 0 = 81\).
 
Now the number \(81\div 7 \) will leave the remainder \(4\).
 
Therefore, \(810\) is not divisible by \(7\).
Divisibility by \(8\): A number is divisible by \(8\) if the number formed by its last three digits is divisible by \(8\).
Example:
1. Let us take the number \(2544\).
 
Rule for \(8\): Last three digits is divisible by \(8\), then the number is divisible by \(8\).
 
The last \(3\) digits are \(544\) and divide it by \(8\).
 
\(544\div8 = 68\).
 
Therefore, \(2544\) is divisible by \(8\).
 
2. Let us take a number \(1260\).
 
Rule for \(8\): Last three digits is divisible by \(8\), then the number is divisible by \(8\).
 
The last \(3\) digits are \(260\) and divide it by \(8\).
 
\(260\div8\) leaves a remainder \(4\).
 
Therefore, \(1260\) is not divisible by \(8\).