Question

find the reminder when 1^2013 + 2^2013 + 3^2013 + ...2012 is devisible by 2013.

Answer 1

Considering the expression, 
1n + 2n + ... + (n - 2)n + (n - 1)n 
Starting at right hand side, expand the bracket, 
(n - 1)n = nn + ... + (-1)n [ This is same as expanding (x+1)2 =x2 +2x+12 we can observe that power of x is decreasing and at the end constant term is obtained]
Now each term (excluding last term) has a n in it (meaning no remainder as when it will be divided by n remainder will be 0) but remember we are excluding the last term.
Since here n = 2013, which is odd, 
(-1)n = -(1n) 
i.e, (n - 1)n = (some multiple of n) - (1n) 
Similarly, the bracket term from R.H.S. , 
(n - 2)n = nn + ... + (-2)n 
(n - 2)n = terms of n + ... - (2n) 
i.e, (n - 2)n = (some multiple of n) - (2n) 
Now compare with the original equation. 
Clearly, the original numbers on the LHS are cancelling with the terms on the RHS because 1n in LHS will be cancelled with 1n in RHS and same for  2 n  and others too.
ie 
{1n + (some multiple of n) - (1n)} + 
{2n + (some multiple of n) - (2n)} + 
....and so on 
Since 2012 is even, these terms match exactly, giving a remainder of 0.


Hope it helps.
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