
Hi Bibs
In earlier case we were grouping the number in 2digit numbers from right so that each group i.e. a 2digit number was multiple of some power of 100 (i.e. 10^{2}) and the groups were alternate +/ with first group from right most side being positive one.
Try to understand the reason on a small 4digit number only.
Say our number is N = abcd = ab*10^{2} + cd
Now we know that 10^{2} when divided by 101 (i.e. 10^{2} + 1) gives remainder 1. So N = {ab(1) + cd} mod 101 = {ab  cd} mod 101
If you had another group of 2digits before ab, then that'd have been multiple of 10^{4} which can be written as 10^{2}*10^{2} = (1)(1) mod 101 = 1 mod 101. So that group would have been positive. Hence the alternate +/ pattern comes in picture here.
Now to enlarge the concept for this question, you can easily observe that in given number the pattern of digits repeat after every 7 digits. So if I consider group of 14 digits collectively, then each group would be multiple of powers of 10^{14}..i.e.
N = 12437681243768*(10^{14})^{49} + 12437681243768*(10^{14})^{48} + 12437681243768*(10^{14})^{47} + ... + 12437681243768*(10^{14})^{1} + 12437681243768
Now just observe that as 10^{2} = 1 mod 101, so 10^{14} = (10^{2})^{7} = (1)^{7} mod 101 = 1 mod 101.
Thus again we are going to get alternate +/ groups of these 14 digits, while dividing with 101, with first group from right most side being positive.
As we have exactly 50 groups of 14 digits each here and half of which are positive and half negative. Thus required remainder is simply zero as all the groups will be cancelled.
Kamal Lohia
