Divisibility Rules on Numbers

A number is divisible by :
2  if the last digit is 0, 2, 4, 6 or 8.
3  if the sum of the digits is divisible by 3.
4  if the number formed by the last two digits is divisible by 4.
5  if the last digit is either 0 or 5.
6  if it is divisible by 2 AND it is divisible by 3.
8  if the number formed by the last three digits is divisible by 8.
9  if the sum of the digits is divisible by 9.
10  if the last digit is 0.
11 if you sum all alternate digits and then subtract this sum from the sum of the rest of the digits answer is 0 or 11.
Example: 1331 ((3+1) - (1+3) = 0). 1331 is divisible by 11.
       25176 ((5+7) - (2+1+6) = 3). 25176 is not divisible by 11.


Divisibility by 7:
This can be tested by a recursive method. A number of the form 10x + y is divisible by 7 if and only if x − 2y is divisible by 7.
In other words, subtract twice the last digit from the number formed by the remaining digits. Continue to do this until a number known to be divisible by 7 is obtained.
The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7.
For example, the number 371: 37 − (2×1) = 37 − 2 = 35; 3 − (2 × 5) = 3 − 10 = −7; thus, since −7 is divisible by 7, 371 is divisible by 7.

Another method is multiplication by 3. A number of the form 10x + y has the same remainder when divided by 7 as 3x + y.
One must multiply the leftmost digit of the original number by 3, add the next digit, take the remainder when divided by 7, and continue from the beginning: multiply by 3, add the next digit, etc.
For example, the number 371: 3×3 + 7 = 16 remainder 2, and 2×3 + 1 = 7. This method can be used to find the remainder of division by 7.