I cud not understand your approach. Kindly solve 4^24/7 for the remainder from your approach and post. Probably you have an easier and shorter way.
Rahul you are trying to compare apples with oranges
When divisor is 5 no matter what is tens digit the units digit always determine the remainder. In case of 7 this does not hold true so approach will be different.
Each number's specific properties regarding their squares, remainders etc are applied after seeing a particular Q and is only used to speed up that particular Q and should not be used to solve every Q of that nature.
Take a look at example below
If Q is whether 249678 is divisible by 11 then I would find sum of each even digit and compare with sum of each odd digit. If sum is equal that means number is divisible by 11. Now if Q is whether same number is divisible by 9 then I will sum each digit and see whether sum is divisible by 9 or not. So different approaches allow me to get the solution faster.
A consistent approach will be to divide the number and see which will give you correct answers for all the similar Qs.
Exactly why I said above :
The "unit digit repeats" method only works because the exercise asks to divide by 5. If it were by 4, 6, 7, etc... it wouldn't work.
That's why I think the "reminder repeats" method is better since it is a general method.
With the "remainder repeats" method, you can solve any of this problems, no need to bother with number divisibility properties (which don't exist for every numbers by the way
But you were right in that if we want to divide by 5, looking at the unit digits is enough (but it is not the case if we divide by some other numbers)
Edit: Plus the question is not "IS Q = 249682 by 11?", but "what is the remainder of Q = 249682 when divided by 11?". When you are finished "finding sum of each even digit and comparing with sum of each odd digit" and when you see the difference is not divisible by 11 all you can say is that Q is not divisible by 11, but it won't give you the remainder