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\(\text{} \)\(\text{Given positive integers $x$ and $y$ such that $x\neq y$ and $\frac{1}{x} + \frac{1}{y} = \frac{1}{12}$, what is the smallest possible value for $x + y$?}\)

 Dec 31, 2019
 #1
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I know that the question was asked here: https://web2.0calc.com/questions/given-positive-integers-x-and-y-such-that-x-not-y, but that is the wrong answer.

 Dec 31, 2019
 #2
avatar+109563 
+1

1/ x   +   1/ y   =   1 / 12

 

[ x + y ]  / xy  =   1 / 12

 

12 [ x + y ]    = xy

 

12x + 12y  = xy

 

12y - xy  =  - 12x

 

y ( 12 - x)  =  - 12x

 

y  =        12x

            _______

              x - 12

 

We  will  have positive integers for x, y   when

 

x         y

13     156

14      84

15      60

16      48

18      36

20      30

24      24         

28      21

 

The smallest  sum   of  x , y    when   x ≠ y   is    49

 

Thanks to EP  for spotting my earlier error   !!!!

 

 

cool cool cool

 Dec 31, 2019
edited by CPhill  Dec 31, 2019
 #3
avatar+109563 
+1

Here's one more way :

Let    x   =    12 + a

Let  y  = 12 +  b

 

1 /  [ 12 + a]   +   1 / [12 + b]    =  1 /12

 

[ 24  + a + b ]  /  [  144 + 12a + 12b +ab ]  =   1/12

 

12* [ 24+ a + b ]   =   [ 144 + 12a + 12b + ab ]

 

288  + 12a + 12b  =  144 + 12a + 12b  + ab

 

144  = ab 

 

So

 

a      b

12   12

16    9

18    8

24    6

36    4

48    3

72    2

144  1

 

The  smallest value  of  x +  y  =  

 

12 + 16 ,  12 +  9   =

 

28  ,    21   =

 

49

 

cool cool cool

 Dec 31, 2019

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