Given a number 12345678901234567890..... up to 500 digits. Find the smallest number n that should be added to the number such that the sum is exactly divisible by 11.
The correct option is C 8
If A is the sum of all the alternate digits of the number starting from the units place and B is the sum of all the alternate digits of the number starting from tens place, then 11's remainder of the number A - B. For one group of ten digits of the number.
A - B = - 5
For the 50 groups, it is - 250
So, when this number is divided by 11, the remainder will be - 8 or +3
Therefore 8 has to be added to the number, so that sum is divisible by 11.