Question

Question #62632

Answer 1

`1^2003+2^2003+3^2003...+2003^2003`

We know, `a^n+b^n " is divisible by "(a+b)` when n is odd because if we put `a=-b` in the expression `a^n+b^n`, it becomes zero.

Following this divisibility rule we can say that each pair of terms of the given sequence having sum of the bases `2004` will be divisible by 2004 eg

`(1^2003+2003^2003); (2^2003+2002^2003);(3^2003+2001^2003);(4^2003+2000^2003);.....1001 "pairs"`

The number of terms being odd the remaining middle term `1002^2003` is to be checked for divisibility by `2004`.

Now `1002^2003/2004=(cancel1002xxcancel1002^501xx1002^2001)/(cancel2xxcancel(1002))`

So the remaining middle term is also divisible by 2004.

Hence we can conclude the given sum of terms is divisible by 2004 leaving no remainder .