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Given positive integers  and  such that  and , what is the smallest possible value for ?\(Given positive integers $x$ and $y$ such that $x\neq y$ and $\frac{1}{x} + \frac{1}{y} = \frac{1}{15}$, what is the smallest possible value for $x + y$?\)

 Jan 29, 2024
 #1
avatar+129742 
+1

1/x  + 1/y =  1/15

 

( x + y)  / (xy)  = 1/15

 

(xy)    / ( x + y)  =  15

 

xy =  15 ( x + y)

 

xy  =15x + 15y

 

xy - 15y =  15x

 

y ( x - 15)  =  15x

 

y =   15x / ( x - 15)

 

x          y

16      240

18       90

20       60

24       40

30       30

40       24

 

Smallest  x + y =  40 + 24  =  64

 

 

cool cool cool

 Jan 29, 2024
 #2
avatar+944 
+1

 

30 + 30 is 60   

.

Bosco  Jan 29, 2024
 #3
avatar+129742 
+1

True....but i'm assuming that x and y are different integers.....

 

However....if  let them  be the same, then your answer is  correct

 

cool cool cool

CPhill  Jan 30, 2024
 #4
avatar+944 
0

 

Chris, you were right all along.  The problem stipulates x not equal to y. 

I had overlooked that in all that run-together text and those dollar signs. 

 

After I realized it, too late to edit, I was relieved that my comment got lost  

in the moderation by AI, which btw I maintain is a little short on the "I" part.  smiley 

Bosco  Feb 1, 2024

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