Okay I think the easiest way here is building the progressions and finding the sum
For example from
10000-19999
10002 10022 10032 10042 10052 10062 10072 10082 10092
10102 10122 10132 10142 10152 10162 10172 10182 10192
difference between 2 rows is exactly 900
so we have 9 numbers in each column and we have to do like 100 rows of combinations so 900 numbers
Sn = ((a1+an)/2) * n
so we have S900 = ((90458 (sum of first row) + 899558)/2) * 900 = 445507200
a900=90458+900 * 899
I tried to count everything this way but I might have made a mistake somewhere I got
4173392237302
Actually I remembered that 11000 is eleven thousand so this method wont apply for 10000-19999 but I think it works for other big numbers
Recounted again and got 4414488098580