Number N is obtained by writing all the digits from 1 to 100 in an

This is the way i solved it:
so, the numbers when in order are these:
123456789
10111213141516171819
20212223242526272829
3031


Since it asks the sum of all of the indivudual numbers, it would be
1+2+3+...+9+1+0+1+1+1+2+1+3+...+2+0+2+1+2+2+...+3+0+3+1

Since the order does not matter in a additive process, we can rewrite above expression as:

1+2+3+...+9+1+1+1+1+...+1+2+3+...+9+2+2+2+...+1+2+3+...+9+3+0+3+1

So, in words, this is a summation of sums of smaller sections

Summation of x from 1 to 9 = 9(10)/2 = 45, and this repeats thrice.
summation of number 1 10 times = 1(10) = 10
summation of number 2 10 times = 2(10) = 20

So, the sum is 45(3)+10+20+3+1+3+0 = 172


and add individual numbers of 172 = 1+7+2 and divide by 9 to get remainder. 1+7+2 = 10/9 = R1