2008 iTest Problems/Problem 41

Revision as of 14:27, 3 July 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 41)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100