I think many of us are aware of divisibility rules but the question is..." why this method?".So let us find out the answer.
1) let's take a three digit no.(rule for 3)
abc=100a+10b+c
= 99a+a+9b+b+c
=(99a+9b)+(a+b+c)
=9(11a+b)+(a+b+c)

now can we say that one part of number is a multiple of 9 and if it is then, number will also be divisible by 3,so we are checking the sum of the digits, which if becomes a multiple of 3 then the complete number also becomes a multiple of 3.


2)lets take a four digit no.(rule for 11)
abcd=1000a+100b+10c+d
=1001a-a+99b+b+11c-c+d
=(1001a+99b+11c)+(b+d)-(a+c)
=11(91a+9b+c)+(b+d)-(a+c)

now can we say that one part of a number is a multiple of 11,so we are checking the difference between the sum of the even and odd places ,if difference comes out to be the multiple of 11 then the entire number is divisible by 11.

3)let us check the divisibility of 7 and 13.(method is same for both 7 and 13).
Take an eight digit number as abcdefgh,now make the sets of three from right to left and then take the difference between the sum of the alternate sets,if the set is not complete put zero before the number.If difference is zero or multiple of 7 or 13 then the number is divisible by 7/13
(fgh+0ab)-(cde)=0 or multiple of 7 or of 13.

CONCLUSION:Any six digit no like(111111,222222,333333........)is always divisible by
7/11/13.For 11 also you can use the same method.


Now the question for you that why we make the sets of three in case of 7/11/13.
post your answers and have a coffee with me.