what will be the remainder when 1223334444...101010(10times) when divisible by 11

4 is the answer
Divisibility rule of 11 is, (sum(odd positioned digits)-sum(even positioned digits))/11
If you observe closely, even numbered digits will become zero as we subtracted. like 22,4444,666666,... of the given number.
For odd numbered digits, we have only one corresponding odd number, 333,55555,...gives only 3,5,.. respectively after subtraction.
Here we have 65digits, as we have to start from right end, we have sum(odd positioned numbers)=25 and sum(even positioned numbers)=10 after applying above two lines.
25-10=15
15/11
remainder is 4.

I think 4 is the remainder...