Actually all the above methods were cool but lenthy(ingleesh prowblem
) i derived something which is really easy and sort if you memorize given numbers....
I found a method which if you keep in mind it will make little faster.....
keep in mind these number
right digit->left digit
4->1(i.e 1 greater than 3 so 4-3=1)
5->2(i.e 2 greater than 3 so 5-3=2)
6->5(i just muggup this number
how you going to memorize these is up to you
now if the unit digit is one of them the right handside digit should be according to that else subtract it.
here is the example
to check 7203
the right hand digit is 3 so the left hand side digit must be 6 but here it is not so subtract 6 from 0 the remaining number(i.e 714) is divisible by 7
let's check it the last digit is now 4 so the previous digit should be 1 and it is now remains 7 which is already divisible by 7
2) to check 509796
the right hand digit is 6 to the left hand side digit must be 5 but it is 9 (oh i forgot to say if the digit is greater than 7 you can also subtract 7 from it so 9 became 2) now subtract 5 from it, it remains 4. now the left hand side number is divisible by 7, let's check
now th number(which is actually 50974) has it's last digit 4 so the previous digit should be 1 but here it's not so subtract 1 from 7 it became 6
now the previous digit should be 5 but it is 9 so subtract 5 from 9
now the number(which is actually 504) has it's last digit 4 so the previous digit should be 1 but here it's again not so
subtract 1 from 0 which gives 49 which is divisible by 7
Actually it's very easy if you remember the Process/Method