What is the remainder when 100^3 + 101^3 + 102^3 + 103^3 + … + 999^3 is divided by 7? I solved it by calculating remainder of (sum of all cubes upto 999) - remainder of (sum of all cubes upto 99). I get 5 as answer. But the correct ans is 0.

Well you can try the following approach: The question follows the sequence as (100cb+101cb+102cb+103cb+104cb+105cb+106cb) gives remainder sequence as: (2cb+3cb+4cb+5cb+6cb+0+1cb)=441/7= rem(0). This cycle of 7 nos. will be repeated again till 999 after 106. Thus, =>106+7x=999 =>7x=893 Now dividing 893 by 7 will give us 4 as remainder. This means that first 4 parts of the remainder sequence will be repeated and rest before them will yield 0 as remainder... Thus, Last 4 sequences i.e. 996cb+997cb+998cb+999cb will yeild remainders as: 2cb+3cb+4cb+5cb=224/7=rem(0).