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Deffense [45]
2 years ago
5

Let $S = 2010 + 2011 + \cdots + 4018$. Compute the residue of $S$, modulo 2009.

Mathematics
1 answer:
Yakvenalex [24]2 years ago
8 0

Notice that

2010 ≡ 1 mod 2009

2011 ≡ 2 mod 2009

2012 ≡ 3 mod 2009

...

4017 ≡ 2008 mod 2009

4018 ≡ 0 mod 2009

So really, <em>S</em> is just the sum of the first 2008 positive integers:

S=\displaystyle\sum_{n=1}^{2008}n=\frac{2008\cdot2009}2

where we invoke the formula

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

and so <em>S</em> ≡ 0 mod 2009.

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