2008 iTest Problems/Problem 41

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Problem

Suppose that $x_1+1=x_2+2=x_3+3=\cdots=x_{2008}+2008=x_1+x_2+x_3+\cdots+x_{2008}+2009$. Find the value of $\left\lfloor|S|\right\rfloor$, where $S=\sum_{n=1}^{2008}x_n$.

Solution

Note that for a given integer $a$, where $1 \le a \le 2008$, \[x_a + a = \sum_{n=1}^{2008}x_n + 2009\] Add up the equations for all $a$ to get \[\sum_{n=1}^{2008}x_n + \frac{2009 \cdot 2008}{2} = 2008(\sum_{n=1}^{2008}x_n + 2009)\] We can substitue $S=\sum_{n=1}^{2008}x_n$ and simplify to make the equation look easier to solve. \[S + 2009 \cdot 1004 = 2008S + 2009 \cdot 2008\] \[-2007S = 2009 \cdot 1004\] \[S = \frac{2009 \cdot 1004}{-2007}\] Thus, $\left\lfloor|S|\right\rfloor = \boxed{1005}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 40
Followed by:
Problem 42
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