2006 SMT/General Problems/Problem 10

Solution

The sum of the first n positive odd integers is $n^2$ . This comes from the fact that $(n+1)^2-n^2 = 2n+1$ (Taking a sum of this equation beginning with $n = 0$ will yield the desired result as the LHS will telescope). Therefore, the sum of the first 2006 odd integers is $2006^2$ . The answer we are looking for is $\sqrt{2006^2} = \boxed{2006}$

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2006 SMT/General Problems/Problem 10

Solution