2006 SMT/General Problems/Problem 10

Solution

First of all, lets note that the sum of all positive integers from $1$ to $n$ inclusive is $\frac{n(n+1)}{2}$ . The sum of all numbers from $1$ to $2006$ is then:

$\frac{2006(2007)}{2} = (1003)(2007)$

Finding the prime factorization of the product, we see that:

$(1003)(2007)=17 \cdot 59 \cdot 3^2 \cdot 223$

Taking the square root, the answer is:

$\sqrt{(1003)(2007)}=\sqrt{17 \cdot 59 \cdot 3^2 \cdot 223} = 3\sqrt{17 \cdot 59 \cdot 223} = \boxed{3\sqrt{223669}}$

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2006 SMT/General Problems/Problem 10

Solution