Respuesta :

LammettHash

Notice that

2010 ≡ 1 mod 2009

2011 ≡ 2 mod 2009

2012 ≡ 3 mod 2009

...

4017 ≡ 2008 mod 2009

4018 ≡ 0 mod 2009

So really, S is just the sum of the first 2008 positive integers:

[tex]S=\displaystyle\sum_{n=1}^{2008}n=\frac{2008\cdot2009}2[/tex]

where we invoke the formula

[tex]\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2[/tex]

and so S ≡ 0 mod 2009.

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