Given positive integers x and y such that $x\neq y$ and $\frac{1}{x} + \frac{1}{y} = \frac{1}{18} + \frac{1}{3}$, what is the smallest possible value for x + y?
+1224 \(\frac{1}{x} + \frac{1}{y} = \frac{1}{18} + \frac{1}{3} = \frac{7}{18}\)
\(\frac{x+y}{xy} = \frac{7}{18}\)
\(18x + 18y = 7xy\)
\(0 = 7xy - 18x - 18y\)
\(324/7 = (7x - 18)(y-18/7)\)
\(324 = (7x - 18)(7y - 18)\)
Create a table with the factors of 324, which is what (7x - 18) and (7y - 18) will multiply to. By brute force, you can find the smallest value of x + y.
