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.

Question

Answer 1

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).

Answer 2

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).

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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.

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.