Difference between revisions of "2006 SMT/General Problems/Problem 10"

@@ Line 1: / Line 1: @@
 ==Solution==
-First of all, lets note that the sum of all positive integers from <math>1</math> to <math>n</math> inclusive is <math>\frac{n(n+1)}{2}</math>. The sum of all numbers from <math>1</math> to <math>2006</math> is then:
-<cmath>\frac{2006(2007)}{2} = (1003)(2007)</cmath>
-Finding the prime factorization of the product, we see that:
-<cmath>(1003)(2007)=17 \cdot 59 \cdot 3^2 \cdot 223</cmath>
-Taking the square root, the answer is:
-<cmath>\sqrt{(1003)(2007)}=\sqrt{17 \cdot 59 \cdot 3^2 \cdot 223} = 3\sqrt{17 \cdot 59 \cdot 223} = \boxed{3\sqrt{223669}}</cmath>

Page