First of all, stop thinking on the number 10001000 and turn your attention to the number 990990 instead. If you solve the problem for 990990 you just have to add 993,995,996993,995,996 & 999999 to it for the final answer. This sum is (a)=3983(a)=3983
Count all the #s divisible by 33: From 33... to 990990 there are 330330 terms. The sum is 330(990+3)/2330(990+3)/2, so (b)=163845(b)=163845
Count all the #s divisible by 55: From 55... to 990990 there are 198198 terms. The sum is 198(990+5)/2198(990+5)/2, so (c)=98505(c)=98505
Now, the GCD (greatest common divisor) of 33 & 55 is 11, so the LCM (least common multiple) should be 3×5=153×5=15.
This means every number that divides by 1515 was counted twice, and it should be done only once. Because of this, you have an extra set of numbers started with 1515 all the way to 990990 that has to be removed from (b)&(c).
Then, from 1515... to 990990 there are 6666 terms and their sum is 66(990+15)/266(990+15)/2, so (d)=33165(d)=33165
The answer for the problem is: (a)+(b)+(c)−(d)=233168(a)+(b)+(c)−(d)=233168
Simple but very fun problem.
