+0

# 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

+1
929
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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
+1015
-6

X=6

Y=6

Mar 22, 2019
#2
+108705
0

That is not correct Nickolas.

Mar 23, 2019
#3
+108705
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$$\frac {1}{x} + \frac {1}{y} = \frac {1}{12}\\~\\ \frac{1}{x}<\frac{1}{12}\qquad \qquad \frac{1}{y}<\frac{1}{12} \\ x>12\qquad \qquad y>12 \\ so\\ x+y>24$$

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Mar 23, 2019
#4
+29252
+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
+108705
+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

Melody  Mar 23, 2019