+209 \( Given\ positive\ integers\ x\ and\ y\ such\ that\ x \not = y\ and\ \frac {1}{x} + \frac {1}{y} = \frac {1}{12},\\ what\ is\ the\ smallest\ possible\ positive\ value\ for\ x + y?\)
I know this is on the website, but, you got 24+24... but x cant be the same as y
+9673 \(\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{12}\\ \dfrac{x+y}{xy} = \dfrac{1}{12}\\ \text{Let }x + y = k, xy = 12k.\\ y(k-y) = 12k\\ -y^2 +ky = 12k\\ y^2 -ky+12k = 0\\ y = \dfrac{k+\sqrt{k^2 -48k}}{2}\\ \implies k^2 - 48k \text{ is a perfect square.}\\ k^2 - 48k \ge 0\\ k^2 - 48k = t^2\\ k \le 0\text{(rejected)} \vee k\ge48\\ k \ge 48\\ k \text{ cannot be 48. If } k = 48\text{, } x = y = 24.\\ k \ge 49\\ k = 49 \text{ satisfies the condition that } k^2 - 48k \text{ is a perfect square.}\\ \min(k) = 49\\ \boxed{\min(x + y) = \color{red}49}\)
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