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Given positive integers x and y such that x doesn't equal y and 1/x+1/y=1/12, what is the smallest possible value for x+y?

 Dec 23, 2018
 #1
avatar+3994 
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We have \(\frac{x+y}{xy}=\frac{1}{12}, 12(x+y)=xy\)\(12x+12y=xy\) . Try to plug in numbers!

 Dec 23, 2018
 #2
avatar+98130 
+2

1/x + 1/y  =  1/12

x + y = xy/12

12(x + y) = xy

12x  - xy   =  -12y

x (12 - y)  = -12y

x ( y - 12)  = 12y

 

x =  12y / ( y - 12)

 

The smallest that y can be and still have x positive is when y = 13

 

(1) If y = 13  x = 156

(2) If y =14  x  = 84

3) If y = 15  x = 60

(4) If y = 16 x = 48

(5) If y = 18 x = 36

(6) If y = 20  x = 30

(7) If y = 24 x = 24      but x and y cannot be the same

(8) If x = 30  y = 20

 

And every other combo will be a repeat  of the sum of x and y for  (1) - (6)

 

So.....the smallest value of x  + y  =   50

 

 

cool cool cool 

 Dec 23, 2018
edited by CPhill  Dec 23, 2018

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