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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 \(x+y\)?

 Mar 22, 2019
 #1
avatar+1014 
-6

X=6

Y=6

 Mar 22, 2019
 #2
avatar+104444 
0

That is not correct Nickolas.

 Mar 23, 2019
 #3
avatar+104444 
+3

\(\frac {1}{x} + \frac {1}{y} = \frac {1}{12}\\~\\ \frac{1}{x}<\frac{1}{12}\qquad \qquad \frac{1}{y}<\frac{1}{12} \\ x>12\qquad \qquad y>12 \\ so\\ x+y>24 \)

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 Mar 23, 2019
 #4
avatar+28159 
+4

Correct Melody! In fact we have:

 

 

 Mar 23, 2019
edited by Alan  Mar 23, 2019
edited by Alan  Mar 23, 2019
edited by Alan  Mar 23, 2019
 #5
avatar+104444 
+1

Thanks Alan,

I did think there might be a higher answer than mine.

I guess I should have said so   :)

 

I also should have thought to use calculus. That never even crossed my mind  sad

Melody  Mar 23, 2019

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