Divisibility rules.

Divisibility by 2 Rule:-
A number is divisible by 2 if its last digit is also (i.e. 0,2,4,6 or 8). Any even number can be divided by 2. Even numbers are multiples of 2. A number is even if ends in 0,2,4,6, or 8.
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Divisibility by 3 Rule:-
A number is divisible by 3 if the sum of its digits is also. Example: 534: 5+3+4=12 and 1+2=3 so 534 is divisible by 3.
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Divisibility by 4 Rule:-
A number is divisible by 4 if the number's  last two digits are divisible by 4.
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Divisibility by 5 Rule:-
A number is divisible by 5 if the its last digit is a 0 or 5.
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Divisibility by 6 Rule:-
Since 6 is a multiple of 2 and 3, the rules for divisibility by 6 are a combination of the rule for 2 and the rule for   3. In other words, a number passes this divisibility test only if it passes the test for 2 and test for 3.
A number is divisible by a composite if it is also divisible by all the prime factors (e.g. is divisible by 6 if divisible by 2 AND by 3; or is divisible by 21 if divisible by 3 AND by 7).
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Divisibility by 7 Rule:-
Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule over and over again as necessary. Example: 826. Twice 6 is 12. So take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also.
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Divisibility by 8 Rule:-
A number passes the test for 8 if the last three digits form a number is divisible 8.
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Divisibility by 9 Rule:-
A number is divisible by 9 if the sum of the digits are evenly divisible 9.
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Divisibility by 10 Rule:-
A number passes the test for 10 if its final digit is 0.
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Divisibility by 11 Rule:-
Subtract the last digit from the remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary.
Example: 19151--> 1915-1 =1914 -->191-4=187 -->18-7=11, so yes, 19151 is divisible by 11.
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Divisibility by 13 Rule:-
Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.
Example: 50661-->5066+4=5070-->507+0=507-->50+28=78 and 78 is 6*13, so 50661 is divisible by 13.
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Divisibility by 17 Rule:-
Subtract five times the last digit from the remaining leading truncated number. If the result is divisible by 17, then so was the first number. Apply this rule over and over again as necessary.
Example: 3978-->397-5*8=357-->35-5*7=0. So 3978 is divisible by 17.
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Divisibility by 19 Rule:-
Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary.
e.g.: 101156-->10115+2*6=10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, so 101156 is divisible by 19.
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Divisibility by 23 Rule:-
Add 7 times the last digit to the remaining leading truncated number. If the result is divisible by 23, then so was the first number. Apply this rule over and over again as necessary.
Example: 17043-->1704+7*3=1725-->172+7*5=207 which is 9*23, so 17043 is also divisible by 23.
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Divisibility by 29 Rule:-
Add three times the last digit to the remaining leading truncated number. If the result is divisible by 29, then so was the first number. Apply this rule over and over again as necessary.
Example: 15689-->1568+3*9=1595-->159+3*5=174-->17+3*4=29, so 15689 is also divisible by 29.
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Divisibility by 31 Rule:-
Subtract three times the last digit from the remaining leading truncated number. If the result is divisible by 31, then so was the first number. Apply this rule over and over again as necessary.
Example: 7998-->799-3*8=775-->77-3*5=62 which is twice 31, so 7998 is also divisible by 31.
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Divisibility by 37 Rule:-
Subtract eleven times the last digit from the remaining leading truncated number. If the result is divisible by 37, then so was the first number. Apply this rule over and over again as necessary.
Example: 23384-->2338-11*4=2294-->229-11*4=185 which is five times 37, so 23384 is also divisible by 37.
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Divisibility by 41 Rule:-
Subtract four times the last digit from the remaining leading truncated number. If the result is divisible by 41, then so was the first number. Apply this rule over and over again as necessary.
Example: 30873-->3087-4*3=3075-->307-4*5=287-->28-4*7=0, remainder is zero and so 30873 is also divisible by 41.
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Divisibility by 43 Rule:-
Add thirteen times the last digit to the remaining leading truncated number. If the result is divisible by 43, then so was the first number. Apply this rule over and over again as necessary.
Example: 3182-->318+13*2=344-->34+13*4=86 which is recognisably twice 43, and so 3182 is also divisible by 43.
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Divisibility by 47 Rule:-
Subtract fourteen times the last digit from the remaining leading truncated number. If the result is divisible by 47, then so was the first number. Apply this rule over and over again as necessary.
Example: 34827-->3482-14*7=3384-->338-14*4=282-->28-14*2=0 , remainder is zero and so 34827 is divisible by 47.
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A COMMON RULE - TO BE UNDERSTOOD:-
We have displayed the recursive divisibility test of number N as f-M*r where f are the front digits of N, r is the rear digit of N and M is some multiplier. And we want to see if N is divisible by some prime P. We need a method to work out the values of M. What you do is to calculate (mentally) the smallest multiple of P which ends in a 9 or a 1. If it's a 9 we are going to ADD, Then we will use the leading digit(s) of the multiple +1 as our multiplier M. If it's a 1 we are going to SUBTRACT later. then we will use the leading digit(s) of the multiple as our multiplier M.

Example for P=17 : three times 17 is 51 which is the smallest multiple of 17 that ends in a 1 or 9. Since it's a 1 we are going to SUBTRACT later. The leading digit is a 5, so we are going to SUBTRACT five times the remainder r. The algorithm was stated above. Now let's do the algebraic proof. Writing N=10f+r, we can multiply by -5 (as shown in the example for 17), getting -5N=-50f-5r. Now we add 51f to both sides (because 51 was the smallest multiple of P=17 to end in a 1 or a 9), giving one f (which we want), so 51f-5N=f-5r. Now if N is divisible by P (here P=17), we can substitute to get 51f-5*17*x=f-5r and rearrange the left side as 17*(3f-5x)=f-5r and therefore f-5r is a multiple of P=17 also.