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Given positive integers $x$ and $y$ such that $x\neq y$ and $\frac{1}{x} + \frac{1}{y} = \frac{1}{18}$, what is the smallest possible value for $x + y$?

 Jul 4, 2021
 #1
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The smallest possible value of x + y = 45, which you get for x = 9 and y = 36.

 Jul 4, 2021
 #2
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45 is incorrect 

Guest Jul 4, 2021
 #3
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\(\frac{1}{x} + \frac{1}{y} = \frac{1}{18}\\ \frac{x+y}{xy} = \frac{1}{18}\\ xy=18x+18y\\ xy-18x-18y=0\\ (x-18)(y-18)=324\)

The pair of numbers that multiplies to 324 that are closest to each other but are not equal to each other are 12 and 27, so the answer is (18+12)+(27+18)

btw a very similar question was already answered here:

https://web2.0calc.com/questions/given-positive-integers-x-and-y-such-that-x-not-to-y-and-1-x-1-y-1-12-what-is-the-smallest-possible-positive-value-for-x-y

textot  Jul 4, 2021

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